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

题意

题目链接

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;
}