洛谷P2881 [USACO07MAR]排名的牛Ranking the Cows(bitset Floyd)

题意

Sol

显然如果题目什么都不说的话需要\(\frac{n * (n - 1)}{2}\)个相对关系

然后求一下传递闭包减掉就行了

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 1001;

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int N, M;

bitset<MAXN> f[MAXN];

int main() {

    N = read(); M = read();

    for(int i = 1; i <= M; i++) {

        int x = read(), y = read();

        f[x][y] = 1;

    }

    for(int k = 1; k <= N; k++)

        for(int i = 1; i <= N; i++)

            if(f[i][k]) f[i] = f[i] | f[k];

    int ans = N * (N - 1) / 2;

    for(int i = 1; i <= N; i++) ans -= f[i].count();

    cout << ans;

    return 0;

}

巴特西

洛谷P2881 [USACO07MAR]排名的牛Ranking the Cows(bitset Floyd)

题意

Sol

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