POJ 3660 Cow Contest 传递闭包确定名次

题目来源：POJ 3660 Cow Contest

题意：n头牛 下面m行 每行x y 代表牛x打败了牛y 问有几头牛的最终排名是确定的

思路：传递闭包 如果x打败了y 令a[x][y]=1 并且a[y][x]=-1 其他不知道的都为0  然后floyd

最后对于每头牛数一下是否有n-1个1或者-1(就是不为0) 如果有n-1个不为0 说明该牛和其他牛都确定了状态

#include <cstdio>#include <algorithm>#include <queue>#include <vector>#include <map>#include <string>using namespace std;const int maxn = 110;int a[maxn][maxn];int n, m;void floyd(){for(int k = 1; k <= n; k++)for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++){if(a[i][k]  == -1  && a[k][j] == -1)a[i][j] = -1;if(a[i][k]  == 1  && a[k][j] == 1)a[i][j] = 1;}}int main(){scanf("%d %d", &n, &m);while(m--){int u, v;scanf("%d %d", &u, &v);a[u][v] = 1;//赢 a[v][u] = -1;//输 }floyd(); int ans = 0;for(int i = 1; i <= n; i++){int cnt = 0;for(int j = 1; j <= n; j++){if(i == j)continue;if(a[i][j])cnt++;}if(cnt == n-1){ans++;//printf("%d\n", i);}}printf("%d\n", ans);return 0;}