floyd

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 110;
int f[N][N];

//核心想法：//i到达j经过前k个城市
//下面表示的是可以确定几个人的排名
//样例：
//5 5  n m 单向
//4 3 u v
//4 2
//3 2
//1 2
//2 5
//可以确定2和5的排名，rank[2] = 4, rank[5] = 5

void solve(){

    int n, m;
    scanf("%d%d", &n, &m);

    int u, v;
    for(int i = 0; i < m; ++i){
        scanf("%d%d", &u, &v);
        f[u][v] = 1;
    }

    for(int i = 1; i <= n; ++i) f[i][i] = 1;

    //i到达j经过前k个城市
    for(int k = 1; k <= n; ++k){
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= n; ++j){
                f[i][j] = f[i][j] | (f[i][k] & f[k][j]);
            }
        }
    }

    int cows = 0;
    for(int i = 1; i <= n; ++i){
        int cnt = 0;
        for(int j = 1; j <= n; ++j){
            cnt += (f[i][j] | f[j][i]);
        }
        //能确定前面有几个rank大于它，后面有几个rank小于它
        if(cnt == n) ++cows;
    }

    printf("%d
", cows);
}

int main(){

    solve();

    return 0;
}