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[kuangbin带你飞]专题七 线段树

时间：2020-11-01 21:18:27      阅读：19      评论：0      收藏：0      [点我收藏+]

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1. HDU1166 敌兵布阵

题目链接

题意：单点更新+区间查询(求和)。

树状数组 (218ms)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

int t, n;
int c[N];
char str[10];

int lowbit(int x) { return x & -x; }

void add(int x, int v) {
    for (; x <= n; x += lowbit(x)) c[x] += v;
}

int sum(int x) {
    int res = 0;
    for (; x; x -= lowbit(x)) res += c[x];
    return res;
}

int main() 
{
    scanf("%d", &t);
    for (int cas = 1; cas <= t; ++cas) {
        scanf("%d", &n);
        int x, y;
        memset(c, 0, sizeof(c));
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &x);
            add(i, x);
        }
        printf("Case %d:\n", cas);
        while (scanf("%s", str) && str[0] != ‘E‘) {
            scanf("%d%d", &x, &y);
            if (str[0] == ‘A‘) add(x, y);
            else if (str[0] == ‘S‘) add(x, -y);
            else printf("%d\n", sum(y) - sum(x - 1));
        }
    }
    return 0;
}

线段树 (234ms)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

struct Node {
    int l, r, sum;
}tree[N<<2];

int t, n;
int a[N];
char str[10];

void push_up(int rt) {
    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    if (l == r) { tree[rt].sum = a[l]; return; }
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
    push_up(rt);
}

void update(int rt, int x, int v) {
    if (tree[rt].l == tree[rt].r) { tree[rt].sum += v; return; }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (x <= mid) update(rt<<1, x, v);
    else update(rt<<1|1, x, v);
    push_up(rt);
}

int query(int rt, int l, int r) {
    if (l <= tree[rt].l && r >= tree[rt].r) return tree[rt].sum;
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    int ans = 0;
    if (l <= mid) ans += query(rt<<1, l, r);
    if (r > mid) ans += query(rt<<1|1, l, r);
    return ans;
}

int main()
{
    scanf("%d", &t);
    for (int cas = 1; cas <= t; ++cas) {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        build(1, 1, n);
        int x, y;
        printf("Case %d:\n", cas);
        while (scanf("%s", str) && str[0] != ‘E‘) {
            scanf("%d%d", &x, &y);
            if (str[0] == ‘A‘) update(1, x, y);
            else if (str[0] == ‘S‘) update(1, x, -y);
            else printf("%d\n", query(1, x, y));
        }
    }
    return 0;
}

zkw(张昆玮) 线段树 (202ms)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

int t, n, m;
int tree[N<<2];
char str[10];

void push_up(int rt) {
    tree[rt] = tree[rt<<1] + tree[rt<<1|1];
} 

void build() {
    for (m = 1; m <= n + 1; m <<= 1);
    for (int i = m + 1; i <= m + n; ++i) scanf("%d", &tree[i]);
    for (int i = m - 1; i; --i) push_up(i);
}

void update(int x, int v) {
    for (x += m; x; x >>= 1) tree[x] += v;
}

int query(int l, int r) {
    int res = 0;
    for (l = l + m - 1, r = r + m + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
        if (~l & 1) res += tree[l ^ 1];
        if (r & 1) res += tree[r ^ 1];
    }
    return res;
}

int main() 
{
    scanf("%d", &t);
    for (int cas = 1; cas <= t; ++cas) {
        scanf("%d", &n);
        build();
        int x, y;
        printf("Case %d:\n", cas);
        while (scanf("%s", str) && str[0] != ‘E‘) {
            scanf("%d%d", &x, &y);
            if (str[0] == ‘A‘) update(x, y);
            else if (str[0] == ‘S‘) update(x, -y);
            else printf("%d\n", query(x, y));
        }
    }
    return 0;
}

2. HDU1754 I Hate It

题目链接

题意：单点更新+区间查询(最值)。

??模板题。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 2e5 + 5;

struct Node {
    int l, r, dat;
}tree[N<<2];

int n, m;
int a[N];
char op[3];

void push_up(int rt) {
    tree[rt].dat = max(tree[rt<<1].dat, tree[rt<<1|1].dat);
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    if (l == r) { tree[rt].dat = a[l]; return; }
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
    push_up(rt);
}

void update(int rt, int x, int v) {
    if (tree[rt].l == tree[rt].r) { tree[rt].dat = v; return; }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (x <= mid) update(rt<<1, x, v);
    else update(rt<<1|1, x, v);
    push_up(rt);
}

int query(int rt, int l, int r) {
    if (l <= tree[rt].l && r >= tree[rt].r) return tree[rt].dat;
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    int res = 0;
    if (l <= mid) res = max(res, query(rt<<1, l, r));
    if (r > mid) res = max(res, query(rt<<1|1, l, r));
    return res;
}

int main() 
{
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        build(1, 1, n);
        int x, y;
        while (m--) {
            scanf("%s%d%d", op, &x, &y);
            if (op[0] == ‘Q‘) printf("%d\n", query(1, x, y));
            else if (op[0] == ‘U‘) update(1, x, y);
        }
    }
    return 0;
}

3. POJ3468 A Simple Problem with Integers

题目链接

题意：区间更新+区间查询(求和)。

??模板题，延迟标记。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 1e5 + 5;

struct Node {
    int l, r;
    ll sum, add;
}tree[N<<2];

int n, q;
int a[N];
char op[3];

void push_down(int rt) {
    if (tree[rt].add) {
        tree[rt<<1].sum += tree[rt].add * (tree[rt<<1].r - tree[rt<<1].l + 1);
        tree[rt<<1|1].sum += tree[rt].add * (tree[rt<<1|1].r - tree[rt<<1|1].l + 1);
        tree[rt<<1].add += tree[rt].add;
        tree[rt<<1|1].add += tree[rt].add;
        tree[rt].add = 0;
    }
}

void push_up(int rt) {
    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    if (l == r) { tree[rt].sum = a[l]; return; }
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
    push_up(rt);
}

void update(int rt, int l, int r, int d) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].sum += (ll)d * (tree[rt].r - tree[rt].l + 1);
        tree[rt].add += d;
        return;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, d);
    if (r > mid) update(rt<<1|1, l, r, d);
    push_up(rt);
}

ll query(int rt, int l, int r) {
    if (l <= tree[rt].l && r >= tree[rt].r) return tree[rt].sum;
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    ll res = 0;
    if (l <= mid) res += query(rt<<1, l, r);
    if (r > mid) res += query(rt<<1|1, l, r);
    return res;
}

int main() 
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    build(1, 1, n);
    while (q--) {
        int x, y, z;
        scanf("%s%d%d", op, &x, &y);
        if (op[0] == ‘C‘) {
            scanf("%d", &z);
            update(1, x, y, z);
        }
        else if (op[0] == ‘Q‘) printf("%lld\n", query(1, x, y));
    }
    return 0;
}

4. POJ2528 Mayor‘s posters

题目链接

题意：市长选举贴高度相同的海报，后面的海报可以贴在前面的上面，最后可以看见的海报。

??离散化用线段树，注意对于如 \([1,10],[1,4],[6,10]\) 等端点重合的样例，离散化将 5 压缩掉，因此需要在距离大于 1 的点中进行插点，不影响最终结果，在 \(2 * N\) 个点中插点，最多 \(4 * N-1\)，因此取 \(N=4e4\)。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 4e4 + 5;

struct Node {
    int l, r, lazytag;
}tree[N<<2];

int cas, n, ans;
int pl[N], pr[N], seg[N<<1];
int vis[N<<1];

void push_down(int rt) {
    if (tree[rt].lazytag) {
        tree[rt<<1].lazytag = tree[rt<<1|1].lazytag = tree[rt].lazytag;
        tree[rt].lazytag = 0;
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].lazytag = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int x) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].lazytag = x;
        return;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, x);
    if (r > mid) update(rt<<1|1, l, r, x);
}

void query(int rt) {
    if (tree[rt].lazytag && !vis[tree[rt].lazytag]) {
        ans++;
        vis[tree[rt].lazytag] = 1;
        return;
    }
    if (tree[rt].l == tree[rt].r) return;
    push_down(rt);
    query(rt<<1);
    query(rt<<1|1);
}

int main() 
{
    scanf("%d", &cas);
    while (cas--) {
        int cnt = 0;
        memset(vis, 0, sizeof(vis));
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d%d", &pl[i], &pr[i]);
            seg[++cnt] = pl[i];
            seg[++cnt] = pr[i];
        }
        sort(seg + 1, seg + cnt + 1);
        int m = unique(seg + 1, seg + cnt + 1) - (seg + 1);
        for (int i = m; i >= 2; --i)
            if (seg[i] - seg[i-1] > 1) seg[++m] = seg[i-1] + 1;
        sort(seg + 1, seg + m + 1);
        build(1, 1, m);
        for (int i = 1; i <= n; ++i) {
            int x = lower_bound(seg + 1, seg + m + 1, pl[i]) - seg;
            int y = lower_bound(seg + 1, seg + m + 1, pr[i]) - seg;
            update(1, x, y, i);
        }
        ans = 0;
        query(1);
        printf("%d\n", ans);
    }
    return 0;
}

5. HDU1698 Just a Hook

题目链接

题意：区间更新+所有数的和。

??延迟标记，最后输出 tree[1].sum 即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 1e5 + 5;

struct Node {
    int l, r, sum, tag;
}tree[N<<2];

int cas, n, q;

void push_up(int rt) {
    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
}

void push_down(int rt) {
    if (tree[rt].tag) {
        tree[rt<<1].tag = tree[rt<<1|1].tag = tree[rt].tag;
        int mid = (tree[rt].l + tree[rt].r) >> 1;
        tree[rt<<1].sum = (mid - tree[rt].l + 1) * tree[rt].tag;
        tree[rt<<1|1].sum = (tree[rt].r - mid) * tree[rt].tag;
        tree[rt].tag = 0;
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].tag = 0;
    if (l == r) { tree[rt].sum = 1; return; }
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
    push_up(rt);
}

void update(int rt, int l, int r, int x) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].sum = x * (tree[rt].r - tree[rt].l + 1);
        tree[rt].tag = x;
        return;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, x);
    if (r > mid) update(rt<<1|1, l, r, x);
    push_up(rt);
}

int main() 
{
    scanf("%d", &cas);
    for (int i = 1; i <= cas; ++i) {
        scanf("%d%d", &n, &q);
        build(1, 1, n);
        int x, y, z;
        while (q--) {
            scanf("%d%d%d", &x, &y, &z);
            update(1, x, y, z);
        }
        printf("Case %d: The total value of the hook is %d.\n", i, tree[1].sum);
    }
    return 0;
}

6. ZOJ1610 Count the Colors

题目链接

题意：区间涂色（可覆盖），求最终可看到几种颜色，以及这些颜色的段数。

??类似于第四题，延迟标记。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 8e3 + 5;

struct Node {
    int l, r, tag;
}tree[N<<2];

int n, last;
int vis[N];

void push_down(int rt) {
    if (~tree[rt].tag) {
        tree[rt<<1].tag = tree[rt<<1|1].tag = tree[rt].tag;
        tree[rt].tag = -1;
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].tag = -1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r); 
}

void update(int rt, int l, int r, int x) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].tag = x;
        return;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, x);
    if (r > mid) update(rt<<1|1, l, r, x);
}

void query(int rt) {
    if (tree[rt].l == tree[rt].r)  {
        if (~tree[rt].tag && tree[rt].tag != last) {
            vis[tree[rt].tag]++;
        }
        last = tree[rt].tag;
        return;
    }
    push_down(rt);
    query(rt<<1);
    query(rt<<1|1);
}

int main() 
{
    while (~scanf("%d", &n)) {
        memset(vis, 0, sizeof(vis));
        build(1, 1, 8000);
        int x, y, z;
        while (n--) {
            scanf("%d%d%d", &x, &y, &z);
            update(1, x + 1, y, z);
        }
        last = -1;
        query(1);
        for (int i = 0; i <= 8000; ++i) {
            if (vis[i]) printf("%d %d\n", i, vis[i]);
        }
        puts("");
    }
    return 0;
}

7. POJ3264 Balanced Lineup

题目链接

题意：农夫 Jorn 的 n 中奶牛排成一列，q 个查询，输出区间最大高度差。

树状数组

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

int n, q;
int a[N], minn[N], maxn[N];

int lowbit(int x) { return x & -x; }

void update(int x, int v) {
    for (; x <= n; x += lowbit(x)) {
        maxn[x] = max(maxn[x], v);
        minn[x] = min(minn[x], v);
    }
}

int ask(int l, int r) {
    int ma = a[l], mi = a[l];
    while (r >= l) {
        ma = max(ma, a[r]); mi = min(mi, a[r]);
        for (r--; r - lowbit(r) >= l; r -= lowbit(r))
            ma = max(ma, maxn[r]), mi = min(mi, minn[r]);
    }
    return ma - mi;
}

int main() 
{
    scanf("%d%d", &n, &q);
    memset(minn, 0x3f, sizeof(minn));
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
        update(i, a[i]);
    }
    int x, y;
    while (q--) {
        scanf("%d%d", &x, &y);
        printf("%d\n", ask(x, y));
    }
    return 0;
}

线段树

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

struct Node {
    int l, r, high, low;
}tree[N<<2];

int n, q;
int a[N];

void push_up(int rt) {
    tree[rt].high = max(tree[rt<<1].high, tree[rt<<1|1].high);
    tree[rt].low = min(tree[rt<<1].low, tree[rt<<1|1].low);
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    if (l == r) {
        tree[rt].high = tree[rt].low = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
    push_up(rt);
}

int query(int rt, int l, int r, int f) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        if (f) return tree[rt].high;
        else return tree[rt].low;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    int res; if (f) res = 0; else res = INF;
    if (l <= mid) {
        if (f) res = max(res, query(rt<<1, l, r, 1));
        else res = min(res, query(rt<<1, l, r, 0));
    }
    if (r > mid) {
        if (f) res = max(res, query(rt<<1|1, l, r, 1));
        else res = min(res, query(rt<<1|1, l, r, 0));
    }
    return res;
}

int main() 
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    build(1, 1, n);
    int x, y;
    while (q--) {
        scanf("%d%d", &x, &y);
        printf("%d\n", query(1, x, y, 1) - query(1, x, y, 0));
    }
    return 0;
}

8. HDU4027 Can you answer these queries?

题目链接

题意：区间修改(开方)+区间查询(求和)。

??每个点更新不一样，只能单点修改，由于 \(2^{64}\) 开 7 次方取整为 1，故还可加限制条件每个点最多更新 7 次，以及区间和等于区间长度时不需继续更新，注意输入数组不开 long long 会 TLE。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 1e5 + 5;

struct Node {
    int l, r;
    ll sum;
}tree[N<<2];

int n, m, cas;
ll a[N];

void push_up(int rt) {
    tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    if (l == r) { tree[rt].sum = a[l]; return; }
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
    push_up(rt);
}

void update(int rt, int l, int r) {
    if (tree[rt].sum == ll(tree[rt].r - tree[rt].l + 1)) return;
    if (tree[rt].l == tree[rt].r) {
        tree[rt].sum = (ll)sqrt(tree[rt].sum);
        return;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r);
    if (r > mid) update(rt<<1|1, l, r);
    push_up(rt);
}

ll query(int rt, int l, int r) {
    if (l <= tree[rt].l && r >= tree[rt].r) return tree[rt].sum;
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    ll res = 0;
    if (l <= mid) res += query(rt<<1, l, r);
    if (r > mid) res += query(rt<<1|1, l, r);
    return res;
}

int main() 
{
    while (~scanf("%d", &n)) {
        for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
        build(1, 1, n);
        scanf("%d", &m);
        printf("Case #%d:\n", ++cas);
        int t, x, y;
        while (m--) {
            scanf("%d%d%d", &t, &x, &y);
            if (x > y) swap(x, y);
            if (t) printf("%lld\n", query(1, x, y));
            else update(1, x, y);
            
        }
        puts("");
    }
    return 0;
}

9. HDU1540 Tunnel Warfare

题目链接

题意：n 个地道，m 个操作，D x: 炸掉第 x 个地道；Q x: 查询第 x 个地道所在的最大区间长度；R x: 重建上一次被炸的地道。

??法一：线段树维护三个信息，ll: 左端点开始最大连续长度，rl: 右端点开始最大连续长度，ml: 区间中最大连续长度。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

struct Node {
    int l, r;
    int ll, rl, ml;
}tree[N<<2];

int n, m;
int sta[N];
char op[3];

void push_up(int rt) {
    tree[rt].ll = tree[rt<<1].ll, tree[rt].rl = tree[rt<<1|1].rl;
    tree[rt].ml = max(tree[rt<<1].ml, tree[rt<<1|1].ml);
    tree[rt].ml = max(tree[rt].ml, tree[rt<<1].rl + tree[rt<<1|1].ll);
    if (tree[rt].ll == tree[rt<<1].r - tree[rt<<1].l + 1) tree[rt].ll += tree[rt<<1|1].ll;
    if (tree[rt].rl == tree[rt<<1|1].r - tree[rt<<1|1].l + 1) tree[rt].rl += tree[rt<<1].rl;
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].ll = tree[rt].rl = tree[rt].ml = r - l + 1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int x, int v) {
    if (tree[rt].l == tree[rt].r) {
        tree[rt].ll = tree[rt].rl = tree[rt].ml = v;
        return;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (x <= mid) update(rt<<1, x, v);
    else update(rt<<1|1, x, v);
    push_up(rt);
}

int query(int rt, int x) {
    if (tree[rt].l == tree[rt].r || !tree[rt].ml || tree[rt].ml == tree[rt].r - tree[rt].l + 1) return tree[rt].ml;
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (x <= mid) {
        if (x >= mid - tree[rt<<1].rl + 1)
            return tree[rt<<1].rl + tree[rt<<1|1].ll;
        else return query(rt<<1, x);
    }
    else {
        if (x <= mid + tree[rt<<1|1].ll)
            return tree[rt<<1|1].ll + tree[rt<<1].rl;
        else return query(rt<<1|1, x);
    }
}

int main() 
{
    while (~scanf("%d%d", &n, &m)) {
        build(1, 1, n);
        int top = 0, x;
        while (m--) {
            scanf("%s", op);
            if (op[0] == ‘D‘) {
                scanf("%d", &x);
                sta[++top] = x;
                update(1, x, 0);
            }
            else if (op[0] == ‘Q‘) {
                scanf("%d", &x);
                printf("%d\n", query(1, x));
            }
            else if (op[0] == ‘R‘) {
                if (!top) continue;
                x = sta[top--];
                update(1, x, 1);
            }
        }
    }
    return 0;
}

??法二：线段树维护区间中被摧毁村庄编号的最值。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

struct Node {
    int l, r;
    int mi, ma;
}tree[N<<2];

int n, m;
int sta[N];
char op[3];

void push_up(int rt) {
    tree[rt].mi = min(tree[rt<<1].mi, tree[rt<<1|1].mi);
    tree[rt].ma = max(tree[rt<<1].ma, tree[rt<<1|1].ma);
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].mi = n + 1, tree[rt].ma = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int x, int v) {
    if (tree[rt].l == tree[rt].r) {
        if (v) { tree[rt].mi = n + 1; tree[rt].ma = 0; }
        else tree[rt].mi = tree[rt].ma = x;
        return;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (x <= mid) update(rt<<1, x, v);
    else update(rt<<1|1, x, v);
    push_up(rt);
}

int query(int rt, int x, int f) {
    int mid = (tree[rt].l + tree[rt].r) >> 1, res;
    if (f) {
        if (tree[rt].r <= x || tree[rt].l == tree[rt].r) {
            return tree[rt].ma;
        }
        res = 0;
        res = max(res, query(rt<<1, x, 1));
        if (mid < x) res = max(res, query(rt<<1|1, x, 1));
        return res;
    }
    else {
        if (tree[rt].l >= x || tree[rt].l == tree[rt].r) {
            return tree[rt].mi;
        }
        res = n + 1;
        res = min(res, query(rt<<1|1, x, 0));
        if (x <= mid) res = min(res, query(rt<<1, x, 0));
        return res;
    }
}

int main() 
{
    while (~scanf("%d%d", &n, &m)) {
        build(1, 1, n);
        int top = 0, x;
        while (m--) {
            scanf("%s", op);
            if (op[0] == ‘D‘) {
                scanf("%d", &x);
                sta[++top] = x;
                update(1, x, 0);
            }
            else if (op[0] == ‘Q‘) {
                scanf("%d", &x);
                int r = query(1, x, 0);
                int l = query(1, x, 1);
                if (l == r) puts("0");
                else printf("%d\n", r - l - 1);
            }
            else if (op[0] == ‘R‘) {
                if (!top) continue;
                x = sta[top--];
                update(1, x, 1);
            }
        }
    }
    return 0;
}

??法三：类似法二的思想，模拟来做。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <set>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

int n, m;
int sta[N];
char op[3];
set<int> s;

int main() 
{
    while (~scanf("%d%d", &n, &m)) {
        s.clear();
        int top = 0, x;
        while (m--) {
            scanf("%s", op);
            if (op[0] == ‘D‘) {
                scanf("%d", &x);
                sta[++top] = x;
                s.insert(x);
            }
            else if (op[0] == ‘Q‘) {
                scanf("%d", &x);
                set<int>::iterator it = s.lower_bound(x);
                int l, r;
                if (it == s.end()) r = n + 1;
                else r = *it;
                if (it == s.begin()) l = 0;
                else l = *(--it);
                if (r == x) puts("0");
                else printf("%d\n", r - l - 1);
            }
            else if (op[0] == ‘R‘) {
                s.erase(sta[top--]);
            }
        }
    }
    return 0;
}

10. HDU3974 Assign the task

题目链接

题意：n 个结点，n-1 个关系，所有结点初始值为 -1，给出 m 个操作，C x: 查询 x 结点的值，T x y: 以 x 为根的子树上所有结点值变成 y。

??dfs 找出树的 dfs 序，找出每个数字左右两边出现的位置，子结点必在两个位置之间，之后用线段树即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

struct Node {
    int l, r, dat, lazy;
}tree[N<<2];

int t, n, m, tot, root;
int head[N], L[N], R[N];
int ver[N], nxt[N];
char op[3];

void add(int x, int y) {
    ver[++tot] = y;
    nxt[tot] = head[x];
    head[x] = tot;
}

void dfs(int x) {
    L[x] = ++tot;
    for (int i = head[x]; i; i = nxt[i]) {
        dfs(ver[i]);
    }
    R[x] = ++tot;	
}

void push_down(int rt) {
    if (tree[rt].lazy) {
        tree[rt<<1].dat = tree[rt<<1|1].dat = tree[rt].lazy;
        tree[rt<<1].lazy = tree[rt<<1|1].lazy = tree[rt].lazy;
        tree[rt].lazy = 0;
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r, tree[rt].lazy = 0;
    if (l == r) { tree[rt].dat = -1; return; }
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int v) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].dat = v;
        tree[rt].lazy = v;
        return;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, v);
    if (r > mid) update(rt<<1|1, l, r, v);
}

int query(int rt, int x) {
    if (tree[rt].l == tree[rt].r) {
        return tree[rt].dat;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (x <= mid) return query(rt<<1, x);
    else return query(rt<<1|1, x);
}

int main() 
{
    scanf("%d", &t);
    for (int cas = 1; cas <= t; ++cas) {
        scanf("%d", &n);
        tot = 0, root = n * (n + 1) / 2;
        for (int i = 1; i <= n; ++i) {
            head[i] = 0;
        }
        int x, y;
        for (int i = 1; i < n; ++i) {
            scanf("%d%d", &x, &y);
            root -= x;
            add(y, x);
        }
        tot = 0;
        dfs(root);
        build(1, 1, n<<1);
        scanf("%d", &m);
        printf("Case #%d:\n", cas);
        while (m--) {
            scanf("%s%d", op, &x);
            if (op[0] == ‘C‘) {
                printf("%d\n", query(1, L[x]));
            }
            else if (op[0] == ‘T‘) {
                scanf("%d", &y);
                update(1, L[x], R[x], y);
            }
        }

    }
    return 0;
}

11. HDU4578 Transformation

题目链接

题意：给定一个长度为 n 的序列，有 4 种操作:

  1. 1 x y c [x, y] 上的值全部加 c
  2. 2 x y c [x, y] 上的值全部乘 c
  3. 3 x y c [x, y] 上的值全部赋为 c
  4. 4 x y p [x, y] 上的值的 p 次方和

??线段树维护延迟标记 加(add), 乘(mul)，以及 sum[i] 表示 i 次方的和，更新时考虑 add、mul 对 sum[i] 的影响

乘法：

  • \(sum[i] = v ^ i * sum[i]\)

加法

  • \(sum[3] = sum[3] + 3v^2 * sum[1] + 3v * sum[2] + len * v^3\)
  • \(sum[2] = sum[2] + 2v * sum[2] + len * v^2\)
  • \(sum[1] = sum[1] + len * v\)

其中 len 为当前区间的长度。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, MOD = 1e4 + 7, N = 1e5 + 5;

struct Node {
    int l, r, add, mul;
    int sum[4];
    void _mul(int v) {
        mul = (mul * v) % MOD;
        add = (add * v) % MOD;
        for (int i = 1; i <= 3; ++i) {
            for (int p = 1; p <= i; ++p) {
                sum[i] = (sum[i] * v) % MOD;
            }
        }
    }
    void _add(int v) {
        add = (add + v) % MOD;
        int len = r - l + 1;
        sum[3] = (sum[3] + 3 * v % MOD * v % MOD * sum[1] % MOD) % MOD;
        sum[3] = (sum[3] + 3 * v % MOD * sum[2] % MOD) % MOD;
        sum[3] = (sum[3] + len * v % MOD * v % MOD * v % MOD) % MOD;
        sum[2] = (sum[2] + 2 * v % MOD * sum[1] % MOD) % MOD;
        sum[2] = (sum[2] + len * v % MOD * v % MOD) % MOD;
        sum[1] = (sum[1] + len * v % MOD) % MOD;
    }
    void calc(int MUL, int ADD) {
        _mul(MUL);
        _add(ADD);
    }
}tree[N<<2];

int n, m;

void push_up(int rt) {
    for (int i = 1; i <= 3; ++i) {
        tree[rt].sum[i] = (tree[rt<<1].sum[i] + tree[rt<<1|1].sum[i]) % MOD;
    }
}

void push_down(int rt) {
    if (tree[rt].add || tree[rt].mul != 1) {
        tree[rt<<1].calc(tree[rt].mul, tree[rt].add);
        tree[rt<<1|1].calc(tree[rt].mul, tree[rt].add);
        tree[rt].add = 0, tree[rt].mul = 1;
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].add = 0, tree[rt].mul = 1;
    for (int i = 1; i <= 3; ++i) tree[rt].sum[i] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int v, int op) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        if (op == 1) tree[rt].calc(1, v);
        else if (op == 2) tree[rt].calc(v, 0);
        else if (op == 3) tree[rt].calc(0, v);
        return;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, v, op);
    if (r > mid) update(rt<<1|1, l, r, v, op);
    push_up(rt);
}

int query(int rt, int l, int r, int p) {
    if (l <= tree[rt].l && r >= tree[rt].r) return tree[rt].sum[p];
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    int res = 0;
    if (l <= mid) res = (res + query(rt<<1, l, r, p)) % MOD;
    if (r > mid) res = (res + query(rt<<1|1, l, r, p)) % MOD;
    return res;
}

int main() 
{
    while (scanf("%d%d", &n, &m), n || m) {
        build(1, 1, n);
        int op, x, y, z;
        while (m--) {
            scanf("%d%d%d%d", &op, &x, &y, &z);
            if (op == 4) {
                printf("%d\n", query(1, x, y, z) % MOD);
            }
            else update(1, x, y, z, op);
        }
    }
    return 0;
}

12. HDU4614 Vases and Flowers

题目链接

题意：有 n 个空花瓶 \(0\sim n-1\)，m 个操作：1 x y: 从 x 位置插 y 多花，有花的花瓶跳过，到最后一个花瓶还有剩余，丢弃剩余花；2 x y: 将区间 \([x,y]\)的花瓶清空。对 1 操作，输出第一个和最后一个插花的位置，如果一朵花都插不了，输出 ‘Can not put any one.‘；对 2 操作，输出区间 \([x,y]\) 内被清空的花瓶数量。

??线段树维护区间空花瓶数 num 和一个延迟标记 tag，对操作 1，判断所求区间空花瓶数大于 0，区间空花瓶递增，可二分找到第一个和最后一个插花位置，将两个位置间 num 置 0；对操作 2，输出总数减空花瓶数，之后区间 num 置为区间长度。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 5e4 + 5;

struct Node {
    int l, r;
    int num, tag;
}tree[N<<2];

int t, n, m;

void push_up(int rt) {
    tree[rt].num = tree[rt<<1].num + tree[rt<<1|1].num;
}

void push_down(int rt) {
    if (~tree[rt].tag) {
        tree[rt<<1].tag = tree[rt<<1|1].tag = tree[rt].tag;
        int mid = (tree[rt].l + tree[rt].r) >> 1;
        tree[rt<<1].num = (mid - tree[rt].l + 1) * tree[rt].tag;
        tree[rt<<1|1].num = (tree[rt].r - mid) * tree[rt].tag;
        tree[rt].tag = -1;
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].num = r - l + 1, tree[rt].tag = -1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int f) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].num = (tree[rt].r - tree[rt].l + 1) * f;
        tree[rt].tag = f;
        return;
    }
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, f);
    if (r > mid) update(rt<<1|1, l, r, f);
    push_up(rt);
}

int query(int rt, int l, int r) {
    if (l <= tree[rt].l && r >= tree[rt].r) return tree[rt].num;
    push_down(rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    int res = 0;
    if (l <= mid) res += query(rt<<1, l, r);
    if (r > mid) res += query(rt<<1|1, l, r);
    return res;
}

int search(int x, int num) {
    int l = x, r = n;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (query(1, x, mid) >= num) r = mid;
        else l = mid + 1;
    }
    return l;
}

int main() 
{
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        build(1, 1, n);
        int op, x, y;
        while (m--) {
            scanf("%d%d%d", &op, &x, &y);
            x++;
            if (op == 1) {
                int cnt = query(1, x, n);
                if (!cnt) puts("Can not put any one.");
                else {
                    int l = search(x, 1);
                    int r = search(x, min(cnt, y));
                    update(1, l, r, 0);
                    printf("%d %d\n", l - 1, r - 1);
                }
            }
            else if (op == 2) {
                y++;
                printf("%d\n", y - x + 1 - query(1, x, y));
                update(1, x, y, 1);
            }
        }
        puts("");
    }
    return 0;
}

13. HDU4553 约会安排

题目链接

题意：长度为 n 的区间，m 个操作：D x: 请求占用长度为 x 的连续区间，返回区间开始位置；N x: 请求占用长度为 x 的连续区间，如果没找到，可以忽略 D 的请求再寻找；S x y: 清空 [x, y] 占用部分。

??建 2 棵线段树维护最长连续区间，分别表示只有女神，以及既可以屌丝又可以女神，函数增加表示线段树的参数，注意输出标点符号是英文的。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 1e5 + 5;

struct Node {
    int l, r;
    int ll, rl, ml, tag;
    inline int length() { return r - l + 1; }
}ds[N<<2], ns[N<<2];

int t, n, m;
char str[10];

void push_up(Node tree[], int rt) {
    tree[rt].ll = tree[rt<<1].ll, tree[rt].rl = tree[rt<<1|1].rl;
    if (tree[rt].ll == tree[rt<<1].length()) tree[rt].ll += tree[rt<<1|1].ll;
    if (tree[rt].rl == tree[rt<<1|1].length()) tree[rt].rl += tree[rt<<1].rl;
    tree[rt].ml = max(tree[rt<<1].rl + tree[rt<<1|1].ll, max(tree[rt<<1].ml, tree[rt<<1|1].ml));
}

void push_down(Node tree[], int rt) {
    if (~tree[rt].tag) {
        tree[rt<<1].tag = tree[rt<<1|1].tag = tree[rt].tag;
        tree[rt<<1].ll = tree[rt<<1].rl = tree[rt<<1].ml = tree[rt].tag ? 0 : tree[rt<<1].length();
        tree[rt<<1|1].ll = tree[rt<<1|1].rl = tree[rt<<1|1].ml = tree[rt].tag ? 0 : tree[rt<<1|1].length();
        tree[rt].tag = -1;
    }
}

void build(Node tree[], int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].ll = tree[rt].rl = tree[rt].ml = r - l + 1;
    tree[rt].tag = -1;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(tree, rt<<1, l, mid);
    build(tree, rt<<1|1, mid + 1, r);
}

void update(Node tree[], int rt, int l, int r, int v) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].tag = v;
        tree[rt].ll = tree[rt].rl = tree[rt].ml = v ? 0 : tree[rt].length();
        return;
    }
    push_down(tree, rt);
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(tree, rt<<1, l, r, v);
    if (r > mid) update(tree, rt<<1|1, l, r, v);
    push_up(tree, rt);
}

int query(Node tree[], int rt, int v) {
    if (tree[rt].ml < v) return 0;
    if (tree[rt].l == tree[rt].r) return tree[rt].l;
    push_down(tree, rt);
    if (tree[rt<<1].ml >= v) return query(tree, rt<<1, v);
    if (tree[rt<<1].rl + tree[rt<<1|1].ll >= v) return tree[rt<<1].r - tree[rt<<1].rl + 1;
    return query(tree, rt<<1|1, v);
}

int main()
{
    scanf("%d", &t);
    for (int cas = 1; cas <= t; ++cas) {
        scanf("%d%d", &n, &m);
        build(ds, 1, 1, n);
        build(ns, 1, 1, n);
        printf("Case %d:\n", cas);
        int x, y;
        while (m--) {
            scanf("%s%d", str, &x);
            if (str[0] == ‘D‘) {
                y = query(ds, 1, x);
                if (y) {
                    printf("%d,let‘s fly\n", y);
                    update(ds, 1, y, y + x - 1, 1);
                }
                else puts("fly with yourself");
            }
            else if (str[0] == ‘N‘) {
                y = query(ds, 1, x);
                if (y) {
                    printf("%d,don‘t put my gezi\n", y);
                    update(ds, 1, y, y + x - 1, 1);
                    update(ns, 1, y, y + x - 1, 1);
                }
                else {
                    y = query(ns, 1, x);
                    if (y) {
                        printf("%d,don‘t put my gezi\n", y);
                        update(ds, 1, y, y + x - 1, 1);
                        update(ns, 1, y, y + x - 1, 1);
                    }
                    else puts("wait for me");
                }
            }
            else if (str[0] == ‘S‘) {
                scanf("%d", &y);
                update(ds, 1, x, y, 0);
                update(ns, 1, x, y, 0);
                puts("I am the hope of chinese chengxuyuan!!");
            }
        }
    }
    return 0;
}

扫描线

接下来几题会涉及到扫描线，建议先学习 OIWiki-扫描线

14. POJ1177 Picture

题目链接

题意：求 n 个矩形的周长并。

??扫描线经典题。线段按 y 升序排序，按照 x 轴建立线段树，线段树维护区间完全覆盖次数 cnt，连续区间个数 num，覆盖长度 len，左右端点是否覆盖 lf、rf。

参考链接

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 2e4 + 5;

struct Line {
    int h, l, r, f;
    Line() {}
    Line(int h, int l, int r, int f) : h(h), l(l), r(r), f(f) {}
    bool operator < (const Line &b) const {
        return h < b.h;
    }
}line[N];

struct Node {
    int l, r, cnt;
    int num, len;
    bool lf, rf;
}tree[N<<2];

int n, ans, pre;
int x[N];
map<int, int> mp;

void push_up(int rt) {
    if (tree[rt].cnt) {
        tree[rt].num = tree[rt].lf = tree[rt].rf = 1;
        tree[rt].len = x[tree[rt].r+1] - x[tree[rt].l];
    }
    else if (tree[rt].l == tree[rt].r) tree[rt].len = tree[rt].num = tree[rt].lf = tree[rt].rf = 0;
    else {
        tree[rt].len = tree[rt<<1].len + tree[rt<<1|1].len;
        tree[rt].num = tree[rt<<1].num + tree[rt<<1|1].num;
        if (tree[rt<<1].rf && tree[rt<<1|1].lf) --tree[rt].num;
        tree[rt].lf = tree[rt<<1].lf;
        tree[rt].rf = tree[rt<<1|1].rf;
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].cnt = tree[rt].num = tree[rt].len = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int k) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].cnt += k;
        push_up(rt);
        return;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, k);
    if (r > mid) update(rt<<1|1, l, r, k);
    push_up(rt);
}

int main() 
{
    scanf("%d", &n);
    int x1, y1, x2, y2;
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        line[i*2-1] = Line(y1, x1, x2, 1);
        line[i*2] = Line(y2, x1, x2, -1);
        x[2*i-1] = x1;
        x[2*i] = x2;
    }
    n <<= 1;
    sort(line + 1, line + n + 1);
    sort(x + 1, x + n + 1);
    int m = unique(x + 1, x + n + 1) - (x + 1);
    for (int i = 1; i <= m; ++i) mp[x[i]] = i;
    build(1, 1, m - 1);
    for (int i = 1; i <= n; ++i) {
        int l = mp[line[i].l], r = mp[line[i].r] - 1;
        update(1, l, r, line[i].f);
        ans += abs(tree[1].len - pre);
        pre = tree[1].len;
        ans += tree[1].num * (line[i+1].h - line[i].h) << 1; // 竖直周长
    }
    printf("%d\n", ans);
    return 0;
}

15. HDU1255 覆盖的面积

题目链接

题意：\(n(1\le n\le1000)\) 个矩形，求覆盖 2 次及以上的矩形面积并。

??稍微修改一下扫描线模板，维护结点被竖直线段完全覆盖次数 cnt，覆盖一次 x 方向长度 len1, 覆盖 2 次及以上 x 方向长度 len2。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 2005;

struct Line {
    double h, l, r;
    int f;
    Line() {}
    Line(double h, double l, double r, int f) : h(h), l(l), r(r), f(f) {}
    bool operator < (const Line &b) const {
        return h < b.h;
    }
}line[N];

struct Node {
    int l, r, cnt;
    double len1, len2;
}tree[N<<2];

int t, n;
double x[N];
map<double, int> mp;

void push_up(int rt) {
    if (tree[rt].cnt > 1) {
        tree[rt].len1 = 0;
        tree[rt].len2 = x[tree[rt].r+1] - x[tree[rt].l];
    }
    else if (tree[rt].cnt == 1) {
        if (tree[rt].l == tree[rt].r) tree[rt].len2 = 0;
        else tree[rt].len2 = tree[rt<<1].len2 + tree[rt<<1].len1 + tree[rt<<1|1].len2 + tree[rt<<1|1].len1;
        tree[rt].len1 = (x[tree[rt].r+1] - x[tree[rt].l]) - tree[rt].len2;
    }
    else {
        if (tree[rt].l == tree[rt].r) tree[rt].len1 = tree[rt].len2 = 0;
        else {
            tree[rt].len1 = tree[rt<<1].len1 + tree[rt<<1|1].len1;
            tree[rt].len2 = tree[rt<<1].len2 + tree[rt<<1|1].len2;
        }
    }
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r, tree[rt].cnt = 0;
    tree[rt].len1 = tree[rt].len2 = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int k) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].cnt += k;
        push_up(rt);
        return;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, k);
    if (r > mid) update(rt<<1|1, l, r, k);
    push_up(rt);
}

int main() 
{
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        double x1, y1, x2, y2;
        for (int i = 1; i <= n; ++i) {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            line[i*2-1] = Line(y1, x1, x2, 1);
            line[i*2] = Line(y2, x1, x2, -1);
            x[i*2-1] = x1;
            x[i*2] = x2;
        }
        n <<= 1;
        sort(line + 1, line + n + 1);
        sort(x + 1, x + n + 1);
        int m = unique(x + 1, x + n + 1) - (x + 1);
        for (int i = 1; i <= m; ++i) mp[x[i]] = i;
        build(1, 1, m - 1);
        double ans = 0;
        for (int i = 1; i <= n; ++i) {
            int l = mp[line[i].l], r = mp[line[i].r] - 1;
            update(1, l, r, line[i].f);
            ans += tree[1].len2 * (line[i+1].h - line[i].h);
        }
        printf("%.2lf\n", ans);
    }
    return 0;
}

16. HDU1542 Atlantis

题目链接

题意：求 \(n(1\le n\le 100)\) 个矩形的面积并。

??看这张别人画的图就明白了

技术图片

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 105;

struct Line {
    double h, l, r;
    int f;
    Line() {}
    Line(double h, double l, double r, int f) : h(h), l(l), r(r), f(f) {}
    bool operator < (const Line &b) const {
        return h < b.h;
    }
}line[N<<1];

struct Node {
    int l, r, cnt;
    double len;
}tree[N<<3];

int n, cas, tot;
double x[N<<1];
map<double, int> mp;

void push_up(int rt) {
    if (tree[rt].cnt) tree[rt].len = x[tree[rt].r+1] - x[tree[rt].l];
    else if (tree[rt].l == tree[rt].r) tree[rt].len = 0;
    else tree[rt].len = tree[rt<<1].len + tree[rt<<1|1].len;
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r;
    tree[rt].cnt = 0, tree[rt].len = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int k) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].cnt += k;
        push_up(rt);
        return;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, k);
    if (r > mid) update(rt<<1|1, l, r, k);
    push_up(rt);
}

int main() 
{
    while (scanf("%d", &n), n) {
        double x1, x2, y1, y2;
        for (int i = 1; i <= n; ++i) {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            line[i*2-1] = Line(y1, x1, x2, 1);
            line[i*2] = Line(y2, x1, x2, -1);
            x[2*i-1] = x1;
            x[2*i] = x2;
        }
        n <<= 1;
        sort(line + 1, line + n + 1);
        sort(x + 1, x + n + 1);
        int m = unique(x + 1, x + n + 1) - (x + 1);
        for (int i = 1; i <= m; ++i) mp[x[i]] = i;
        build(1, 1, m - 1);
        double ans = 0;
        for (int i = 1; i <= n; ++i) {
            int l = mp[line[i].l], r = mp[line[i].r] - 1;
            update(1, l, r, line[i].f);
            ans += tree[1].len * (line[i+1].h - line[i].h);
        }
        printf("Test case #%d\nTotal explored area: %.2lf\n\n", ++cas, ans);
    }
    return 0;
}

17. HDU3642 Get The Treasury

题目链接

题意：\(n(1\le n\le1000)\) 个长方体，求相交 3 次及以上部分的体积。

??\(z\in[-500,500]\)，枚举 \(z_i\) 在 \([z_i,z_{i+1})\) 用扫描线转化为二维。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 2005;

struct Plane {
    int y, x1, z1, x2, z2, f;
    Plane() {}
    Plane(int y, int x1, int z1, int x2, int z2, int f) : y(y), x1(x1), z1(z1), x2(x2), z2(z2), f(f) {}
    bool operator < (const Plane &b) const {
        return y < b.y;
    }
}plane[N], tmp[N];

struct Node {
    int l, r, cnt;
    int len1, len2, len3;
}tree[N<<2];

int t, n;
int x[N], z[N];
map<int, int> mp;

void push_up(int rt) {
    if (tree[rt].cnt) tree[rt].len1 = x[tree[rt].r+1] - x[tree[rt].l];
    else if (tree[rt].l == tree[rt].r) tree[rt].len1 = 0;
    else tree[rt].len1 = tree[rt<<1].len1 + tree[rt<<1|1].len1;

    if (tree[rt].cnt >= 2) tree[rt].len2 = x[tree[rt].r+1] - x[tree[rt].l];
    else if (tree[rt].l == tree[rt].r) tree[rt].len2 = 0;
    else if (tree[rt].cnt == 1) tree[rt].len2 = tree[rt<<1].len1 + tree[rt<<1|1].len1;
    else if (!tree[rt].cnt) tree[rt].len2 = tree[rt<<1].len2 + tree[rt<<1|1].len2;

    if (tree[rt].cnt >= 3) tree[rt].len3 = x[tree[rt].r+1] - x[tree[rt].l];
    else if (tree[rt].l == tree[rt].r) tree[rt].len3 = 0;
    else if (tree[rt].cnt == 2) tree[rt].len3 = tree[rt<<1].len1 + tree[rt<<1|1].len1;
    else if (tree[rt].cnt == 1) tree[rt].len3 = tree[rt<<1].len2 + tree[rt<<1|1].len2;
    else if (!tree[rt].cnt) tree[rt].len3 = tree[rt<<1].len3 + tree[rt<<1|1].len3;
}

void build(int rt, int l, int r) {
    tree[rt].l = l, tree[rt].r = r, tree[rt].cnt = 0;
    tree[rt].len1 = tree[rt].len2 = tree[rt].len3 = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid + 1, r);
}

void update(int rt, int l, int r, int k) {
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].cnt += k;
        push_up(rt);
        return;
    }
    int mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) update(rt<<1, l, r, k);
    if (r > mid) update(rt<<1|1, l, r, k);
    push_up(rt);
}

int main() 
{
    scanf("%d", &t);
    for (int cas = 1; cas <= t; ++cas) {
        scanf("%d", &n);
        int x1, x2, y1, y2, z1, z2;
        for (int i = 1; i <= n; ++i) {
            scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);
            plane[i*2-1] = Plane(y1, x1, z1, x2, z2, 1);
            plane[i*2] = Plane(y2, x1, z1, x2, z2, -1);
            x[i*2-1] = x1;
            x[i*2] = x2;
            z[i*2-1] = z1;
            z[i*2] = z2;
        }
        n <<= 1;
        sort(x + 1, x + n + 1);
        sort(z + 1, z + n + 1);
        int m1 = unique(x + 1, x + n + 1) - (x + 1);
        int m2 = unique(z + 1, z + n + 1) - (z + 1);
        for (int i = 1; i <= m1; ++i) mp[x[i]] = i;
        n >>= 1;
        ll ans = 0;
        for (int i = 1; i < m2; ++i) {
            build(1, 1, m1 - 1);
            int tot = 0;
            for (int j = 1; j <= n; ++j) {
                if (plane[j*2-1].z1 <= z[i] && plane[j*2-1].z2 > z[i]) {
                    tmp[++tot] = plane[j*2-1];
                    tmp[++tot] = plane[j*2];
                }
            }
            sort(tmp + 1, tmp + tot + 1);
            ll res = 0;
            for (int j = 1; j < tot; ++j) {
                int l = mp[tmp[j].x1], r = mp[tmp[j].x2] - 1;
                update(1, l, r, tmp[j].f);
                res += (ll)tree[1].len3 * (tmp[j+1].y - tmp[j].y);
            }
            ans += res * (z[i+1] - z[i]);
        }
        printf("Case %d: %lld\n", cas, ans);
    }
    return 0;
}

[kuangbin带你飞]专题七 线段树

标签：初始   延迟标记   最大连续   之间   tin   端点   add   geo   algo   

原文地址：https://www.cnblogs.com/2inf/p/13868482.html
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