码迷，mamicode.com
首页 > 其他好文 > 详细

NOIP2016题目整合

时间：2016-11-29 21:36:48      阅读：192      评论：0      收藏：0      [点我收藏+]

标签：深度   状压   point   scan   lib   pac   utc   struct   build   

今天终于拿到官方数据，兴致勃勃地全 A 了。

Day 1 T1 toy

处理一下正负号加加减减取模乱搞就好了。

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

int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
    return x * f;
}

#define maxn 100010
int n, q;
struct Toy { int tp; char S[15]; } ts[maxn];

int main() {
	freopen("toy.in", "r", stdin);
	freopen("toy.out", "w", stdout);
    n = read(); q = read();
    for(int i = 0; i < n; i++) {
        int t = read(); scanf("%s", ts[i].S);
        if(!t) ts[i].tp = 1;
        else ts[i].tp = -1;
    }
    int p = 0;
    while(q--) {
        int a = read(), b = read();
        if(a) a = 1; else a = -1;
        p += a * ts[p].tp * b;
        p = (p % n + n) % n;
    }
    
    printf("%s\n", ts[p].S);
    
    return 0;
}

Day 1 T2 running

受到“链”和“S 恒为 1”和“T 恒为 1”的特殊点的启发，我们发现可以将每条链 [Si, Ti] 分成 [Si, Ci] 和 [Ci, Ti] 两条，然后对于一个全部可观测到的链 [Ci, Ti] 将所有 Wi 减去深度是一个定值，对于 [Si, Ci] 的部分所有 Wi 加上深度是一个定值。于是树链剖分再打打标记统计就好了。

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

int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
    return x * f;
}

#define maxn 300010
#define maxm 600010
#define maxs 12000010
int n, q, m, head[maxn], next[maxm], to[maxm], W[maxn];
struct Player {
    int s, t, c;
    Player() {}
    Player(int _, int __): s(_), t(__) {}
} ps[maxn];

void AddEdge(int a, int b) {
    to[++m] = b; next[m] = head[a]; head[a] = m;
    swap(a, b);
    to[++m] = b; next[m] = head[a]; head[a] = m;
    return ;
}

int fa[maxn], dep[maxn], siz[maxn], son[maxn], top[maxn], pos[maxn], pid[maxn], clo;
void build(int u) {
    siz[u] = 1;
    for(int e = head[u]; e; e = next[e]) if(to[e] != fa[u]) {
        fa[to[e]] = u;
        dep[to[e]] = dep[u] + 1;
        build(to[e]);
        siz[u] += siz[to[e]];
        if(siz[son[u]] < siz[to[e]]) son[u] = to[e];
    }
    return ;
}
void gett(int u, int tp) {
    top[u] = tp; pid[++clo] = u; pos[u] = clo;
    if(son[u]) gett(son[u], tp);
    for(int e = head[u]; e; e = next[e])
        if(to[e] != fa[u] && to[e] != son[u]) gett(to[e], to[e]);
    return ;
}
int lca(int a, int b) {
    int f1 = top[a], f2 = top[b];
    while(f1 != f2) {
        if(dep[f1] < dep[f2]) swap(f1, f2), swap(a, b);
        a = fa[f1]; f1 = top[a];
    }
    return dep[a] < dep[b] ? a : b;
}

struct Info {
    int c, fir[maxn], aft[maxs], val[maxs];
    void clear() {
        c = 0;
        memset(fir, 0, sizeof(fir));
        return ;
    }
    void AddInfo(int x, int v) {
        val[++c] = v; aft[c] = fir[x]; fir[x] = c;
        return ;
    }
} add, del;
int tot[maxn<<1], ans[maxn];
void process(int x, int t, int a, int v) {
    int f = top[a];
    while(f != top[t]) {
        add.AddInfo(pos[f], v);
        del.AddInfo(pos[a], v);
        a = fa[f]; f = top[a];
    }
    add.AddInfo(pos[t], v);
    del.AddInfo(pos[a], v);
    return ;
}

int main() {
	freopen("running.in", "r", stdin);
	freopen("running.out", "w", stdout);
    n = read(); q = read();
    for(int i = 1; i < n; i++) {
        int a = read(), b = read();
        AddEdge(a, b);
    }
    for(int i = 1; i <= n; i++) W[i] = read();
    for(int i = 1; i <= q; i++) {
        int s = read(), t = read();
        ps[i] = Player(s, t);
    }
    build(1);
    gett(1, 1);
    
    for(int i = 1; i <= n; i++) W[i] -= dep[i];
    add.clear(); del.clear(); memset(tot, 0, sizeof(tot));
    for(int i = 1; i <= q; i++) {
        ps[i].c = lca(ps[i].s, ps[i].t);
        process(i, ps[i].c, ps[i].t, dep[ps[i].s] - dep[ps[i].c] - dep[ps[i].c]);
    }
    for(int i = 1; i <= n; i++) {
        for(int e = add.fir[i]; e; e = add.aft[e]) {
            int v = add.val[e] + n;
            tot[v]++;
        }
        int u = pid[i];
        ans[u] += tot[W[u]+n];
        for(int e = del.fir[i]; e; e = del.aft[e]) {
            int v = del.val[e] + n;
            tot[v]--;
        }
    }
    
    for(int i = 1; i <= n; i++) W[i] += (dep[i] << 1);
    add.clear(); del.clear(); memset(tot, 0, sizeof(tot));
    for(int i = 1; i <= q; i++)
        process(i, ps[i].c, ps[i].s, dep[ps[i].s]);
    for(int i = 1; i <= n; i++) {
        for(int e = add.fir[i]; e; e = add.aft[e]) {
            int v = add.val[e];
            tot[v]++;
        }
        int u = pid[i];
        ans[u] += tot[W[u]];
        for(int e = del.fir[i]; e; e = del.aft[e]) {
            int v = del.val[e];
            tot[v]--;
        }
    }
    
    for(int i = 1; i <= q; i++)
        if(W[ps[i].c] == dep[ps[i].s]) ans[ps[i].c]--;
    
    for(int i = 1; i <= n; i++) printf("%d%c", ans[i], i < n ? ‘ ‘ : ‘\n‘);
    
    return 0;
}

Day 1 T3 classroom

这题我考场上想到正解但是因为邻接矩阵连边时忘记取 min 就 sb 了。。。。。

设 f[0][i][j] 表示前 i 条请求使用了 j 条，最后一条没取的最小期望距离；f[1][i][j] 表示前 i 条请求使用了 j 条，最后一条取了的最小期望距离。然后因为每次走哪条边是独立的，转移时直接乘上概率累加上就好了。

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

int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
    return x * f;
}

#define maxn 2010
#define maxv 310
#define oo 1000000000
int n, m, v, e, c[maxn], d[maxn], D[maxn][maxn];
double p[maxn], f[2][maxn][maxn];

void up(double& a, double b) {
    a = min(a, b);
    return ;
}

int main() {
	freopen("classroom.in", "r", stdin);
	freopen("classroom.out", "w", stdout);
    n = read(); m = read(); v = read(); e = read();
    for(int i = 1; i <= n; i++) c[i] = read();
    for(int i = 1; i <= n; i++) d[i] = read();
    for(int i = 1; i <= n; i++) scanf("%lf", &p[i]);
    for(int i = 1; i <= v; i++) {
        D[i][i] = 0;
        for(int j = i + 1; j <= v; j++)
            D[i][j] = D[j][i] = oo;
    }
    for(int i = 1; i <= e; i++) {
        int a = read(), b = read(), c = read();
        D[a][b] = min(D[a][b], c);
        D[b][a] = min(D[b][a], c);
    }
    for(int k = 1; k <= v; k++)
        for(int i = 1; i <= v; i++)
            for(int j = 1; j <= v; j++)
                D[i][j] = min(D[i][j], D[i][k] + D[k][j]);
    
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= m; j++) f[0][i][j] = f[1][i][j] = 1e9;
    f[0][0][0] = 0.0;
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= min(i + 1, m); j++) {
            double P = p[i], np = p[i+1], p1 = 1.0 - P, np1 = 1.0 - np;
            up(f[0][i+1][j], f[0][i][j] + D[c[i]][c[i+1]]);
            up(f[1][i+1][j+1], f[0][i][j] + np * D[c[i]][d[i+1]] + np1 * D[c[i]][c[i+1]]);
            up(f[0][i+1][j], f[1][i][j] + P * D[d[i]][c[i+1]] + p1 * D[c[i]][c[i+1]]);
            up(f[1][i+1][j+1], f[1][i][j] + P * np * D[d[i]][d[i+1]] + P * np1 * D[d[i]][c[i+1]] + p1 * np * D[c[i]][d[i+1]] + p1 * np1 * D[c[i]][c[i+1]]);
        }
    
    double ans = 1e9;
    for(int i = 0; i <= m; i++) up(ans, min(f[0][n][i], f[1][n][i]));
    printf("%.2lf\n", ans);
    
    return 0;
}

Day 2 T1 problem

用递推法求组合数，实时模 k。

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

int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
    return x * f;
}

#define maxn 2010
int C[2][maxn], f[maxn][maxn];

int main() {
	freopen("problem.in", "r", stdin);
	freopen("problem.out", "w", stdout);
    int size = 2000;
    bool cur = 0;
    int T = read(), k = read();
    for(int i = 0; i <= size; i++, cur ^= 1) {
        C[cur][0] = 1; C[cur][i] = 1;
        for(int j = 1; j < i; j++) C[cur][j] = (C[cur^1][j-1] + C[cur^1][j]) % k;
        for(int j = 0; j <= i; j++) {
            f[i][j] = (!C[cur][j]);
            int t = 0;
            if(j) f[i][j] += f[i][j-1], t++;
            if(j <= i - 1) f[i][j] += f[i-1][j], t++;
            if(t == 2 && i && j) f[i][j] -= f[i-1][j-1];
        }
    }
    while(T--) {
        int n = read(), m = read();
        printf("%d\n", f[n][min(n,m)]);
    }
    
    return 0;
}

Day 2 T2 earthworm

受到 q = 0 即蚯蚓长度没有增加的数据启发，我们发现每次坎出的两条蚯蚓的长度一定是单调降的，于是可以 O(n) 做了。

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

int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
    return x * f;
}

#define maxn 100010
#define maxm 7100010
#define LL long long
int n, m, q, u, v, t, A[maxn], B[maxm], C[maxm], lb, rb, lc, rc, ans[maxm], cnt;

bool cmp(int a, int b) { return a > b; }

void process(int tmp, int i, int len) {
    int nb = (LL)tmp * u / v, nc = tmp - nb;
    B[++rb] = nb - len - q;
    C[++rc] = nc - len - q;
    if(i % t == 0) ans[++cnt] = tmp;
    return ;
}

int main() {
	freopen("earthworm.in", "r", stdin);
	freopen("earthworm.out", "w", stdout);
    n = read(); m = read(); q = read(); u = read(); v = read(); t = read();
    for(int i = 1; i <= n; i++) A[i] = read();
    sort(A + 1, A + n + 1, cmp);
//    for(int i = 1; i <= n; i++) printf("%d%c", A[i], i < n ? ‘ ‘ : ‘\n‘);
    
    lb = 1; rb = 0; lc = 1; rc = 0;
    int pa = 1, len = 0;
    for(int i = 1; i <= m; i++, len += q) {
        int a, b, c;
        a = (pa <= n) ? A[pa] + len : -1;
        b = (lb <= rb) ? B[lb] + len : -1;
        c = (lc <= rc) ? C[lc] + len : -1;
        if(a >= b && a >= c) {
            pa++;
            process(a, i, len);
        }
        else if(b >= a && b >= c) {
            lb++;
            process(b, i, len);
        }
        else {
            lc++;
            process(c, i, len);
        }
    }
    for(int i = 1; i <= cnt; i++) printf("%d%c", ans[i], i < cnt ? ‘ ‘ : ‘\n‘);
    if(!cnt) putchar(‘\n‘);
    cnt = 0;
    for(int i = 1; i <= n + m; i++) {
        int a, b, c;
        a = (pa <= n) ? A[pa] + len : -1;
        b = (lb <= rb) ? B[lb] + len : -1;
        c = (lc <= rc) ? C[lc] + len : -1;
        if(a >= b && a >= c) {
            pa++;
            if(i % t == 0) ans[++cnt] = a;
        }
        else if(b >= a && b >= c) {
            lb++;
            if(i % t == 0) ans[++cnt] = b;
        }
        else {
            lc++;
            if(i % t == 0) ans[++cnt] = c;
        }
    }
    for(int i = 1; i <= cnt; i++) printf("%d%c", ans[i], i < cnt ? ‘ ‘ : ‘\n‘);
    if(!cnt) putchar(‘\n‘);
    
    return 0;
}

Day 2 T3 angrybirds

状压 dp，设 f[S] 表示干掉集合 S 的猪需要的最少抛物线条数，转移时找到第一个没有被干掉的猪，再枚举另一头猪，两点确定一条过 (0, 0) 的抛物线，然后转移到当前集合与抛物线经过猪 的集合的并集。注意到每条抛物线经过猪的集合是可以预处理的。

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

int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
    return x * f;
}

#define maxn 23
#define maxs 362154
const double eps = 1e-6;
struct Point {
    double x, y;
    Point() {}
    Point(double _, double __): x(_), y(__) {}
    bool operator < (const Point& t) const { return x != t.x ? x < t.x : y < t.y; }
} ps[maxn];
int f[maxs], ls[maxn][maxn];

bool on(double a, double b, Point p) {
    return fabs(a * p.x * p.x + b * p.x - p.y) <= eps;
}

void up(int& a, int b) {
    if(a < 0) a = b;
    else a = min(a, b);
    return ;
}

int main() {
	freopen("angrybirds.in", "r", stdin);
	freopen("angrybirds.out", "w", stdout);
    int T = read();
    while(T--) {
        int n = read(); read();
        for(int i = 0; i < n; i++) scanf("%lf%lf", &ps[i].x, &ps[i].y);
        sort(ps, ps + n);
        memset(f, -1, sizeof(f));
        memset(ls, 0, sizeof(ls));
        f[0] = 0;
        for(int i = 0; i < n; i++)
        	for(int j = 0; j < n; j++) if(i != j) {
                double x1 = ps[i].x, y1 = ps[i].y, x2 = ps[j].x, y2 = ps[j].y;
                double b = (x1 * x1 * y2 - x2 * x2 * y1) / (x1 * x1 * x2 - x2 * x2 * x1);
                double a = (y1 - b * x1) / (x1 * x1);
                if(a >= 0.0) continue;
                int S = 0;
                for(int k = 0; k < n; k++) if(((S >> k & 1) ^ 1) && on(a, b, ps[k]))
                    S |= (1 << k);
        		ls[i][j] = S;
        	}
        int all = (1 << n) - 1;
        for(int S = 0; S <= all; S++) if(f[S] >= 0)
            for(int j = 0; j < n; j++) if((S >> j & 1) ^ 1) {
                int tS = S | (1 << j);
                up(f[tS], f[S] + 1);
                for(int i = j + 1; i < n; i++) if((tS >> i & 1) ^ 1)
                    up(f[tS | ls[i][j]], f[S] + 1);
                break;
            }
        printf("%d\n", f[all]);
    }
    
    return 0;
}

这次 NOIP 为什么都是第二题最难 TAT

NOIP2016题目整合

标签：深度   状压   point   scan   lib   pac   utc   struct   build   

原文地址：http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6114798.html
踩
(0)
赞
(0)
   
举报
评论 一句话评论（0）
登录后才能评论！
分享档案
更多>
2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)
周排行
mamicode.com排行更多图片
更多
友情链接
兰亭集智   国之画   百度统计   站长统计   阿里云   chrome插件   新版天听网
关于我们 - 联系我们 - 留言反馈
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码！