ural 1707. Hypnotoad's Secret(线段树)

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题目链接：ural 1707. Hypnotoad‘s Secret

题目大意：给定N和M，然后N组s0, t0, Δs, Δt, k，每组可以计算出k个星星的坐标；M组a0, b0, c0, d0, Δa, Δb, Δc, 

Δd, q，每组要求算出q个矩形，判断矩形内是否包含星星，对于q≥20的情况要根据公式计算一个值即可。

解题思路：计算出所有的星星坐标和矩阵，这个每的说了，将矩阵差分成两点，通过计算出每个点左下角有多少个星

星，然后用容斥计算出矩阵内是否有点。这个属于线段树的一个应用。

#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;
typedef long long ll;

const ll mod = 200904040963LL;
const int maxn = 300000;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], s[maxn << 2];

inline void pushup(int u) {
    s[u] = s[lson(u)] + s[rson(u)];
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    s[u] = 0;

    if (l == r)
        return;

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify(int u, int x) {
    if (lc[u] == x && x == rc[u]) {
        s[u] += 1;
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (x <= mid)
        modify(lson(u), x);
    else
        modify(rson(u), x);
    pushup(u);
}

int query(int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r)
        return s[u];

    int mid = (lc[u] + rc[u]) / 2, ret = 0;
    if (l <= mid)
        ret += query(lson(u), l, r);
    if (r > mid)
        ret += query(rson(u), l, r);
    return ret;
}

struct point {
    int t,  x, y;
    point (int x = 0, int y = 0, int t = 0) {
        set(x, y);
        this->t = t;
    }
    void set (int x, int y) {
        this->x = x;
        this->y = y;
    }
    friend bool operator < (const point& a, const point& b) {
        if (a.x != b.x)
            return a.x < b.x;
        if (a.y != b.y)
            return a.y < b.y;
        return a.t < b.t;
    }
};

struct state {
    point a, b;
    state (point a, point b) {
        this->a = a;
        this->b = b;
    }
};

int N, M, P;
ll T[350];
vector<int> pos, tmp, cnt;
vector<point> vec;
vector<state> que;
map<point, int> G;

inline void add_point (ll x, ll y, int k) {
    vec.push_back(point(x, y, k));
    tmp.push_back(y);
}

inline void add_state (int a, int b, int c, int d) {
    add_point(a - 1, b - 1, 1);
    add_point(c, d, 1);

    int u = vec.size();
    que.push_back(state(vec[u-2], vec[u-1]));

    add_point(a - 1, d, 1);
    add_point(c, b - 1, 1);
}

inline int find (int a) {
    return lower_bound(pos.begin(), pos.end(), a) - pos.begin();
}

void init () {
    G.clear();
    cnt.clear();
    vec.clear();
    que.clear();
    pos.clear();
    tmp.clear();

    int a, b, c, d, aa, bb, cc, dd, k;

    for (int i = 0; i < M; i++) {
        scanf("%d%d%d%d%d", &a, &b, &aa, &bb, &k);
        a = (a + N) % N; b = (b + N) % N;
        for (int j = 0; j < k; j++) {
            add_point(a, b, 0);
            a = (a + aa + N) % N;
            b = (b + bb + N) % N;
        }
    }

    scanf("%d", &P);
    for (int i = 0; i < P; i++) {
        scanf("%d%d%d%d", &a, &b, &c, &d);
        scanf("%d%d%d%d%d", &aa, &bb, &cc, &dd, &k);
        a = (a + N) % N; b = (b + N) % N;
        c = (c + N) % N; d = (d + N) % N;

        cnt.push_back(k);
        for (int j = 0; j < k; j++) {
            add_state(min(a, b), min(c, d), max(a, b), max(c, d));

            a = (a + aa + N) % N; b = (b + bb + N) % N;
            c = (c + cc + N) % N; d = (d + dd + N) % N;
        }
    }

    sort(vec.begin(), vec.end());
    sort(tmp.begin(), tmp.end());
    pos.push_back(tmp[0]);
    for (int i = 1; i < tmp.size(); i++) {
        if (tmp[i] != tmp[i-1])
            pos.push_back(tmp[i]);
    }
    build(1, 0, pos.size());
}

inline int judge (state u) {
    return G[u.b] + G[u.a] - G[point(u.b.x, u.a.y, 1)] - G[point(u.a.x, u.b.y, 1)];
}

void solve () {
    for (int i = 0; i < vec.size(); i++) {
        if (vec[i].t)
            G[vec[i]] = query(1, 0, find(vec[i].y));
        else
            modify(1, find(vec[i].y));
    }

    int mv = 0;
    for (int i = 0; i < cnt.size(); i++) {
        if (cnt[i] > 20) {
            ll ret = 0;
            for (int j = 0; j < cnt[i]; j++)
                ret = (ret + (judge(que[mv + j]) > 0 ? 1 : 0) * T[j]) % mod;
            printf("%lld", ret);
        } else {
            for (int j = 0; j < cnt[i]; j++)
                printf("%d", judge(que[mv + j]) > 0 ? 1 : 0);
        }

        mv += cnt[i];
        printf("\n");
    }
}

int main () {

    T[0] = 1;
    for (int i = 1; i <= 345; i++)
        T[i] = T[i-1] * 7 % mod;

    while (scanf("%d%d", &N, &M) == 2) {
        init();
        solve();
    }
    return 0;
}

ural 1707. Hypnotoad's Secret(线段树)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/40053193