Floyd_Warshall POJ 3660 Cow Contest

题目传送门

题意: m组关系,A能打败B,问最后有几头牛的排名能确定

分析:如果排名确定,那么能打败它的到它一定通,它到能打败的一定能通,也就是和为n-1.用Floyd的传递闭包

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 1e2 + 5;
bool d[N][N];
int in[N];
int n, m;

void Floyd_Warshall(void)	{
	for (int k=1; k<=n; ++k)	{
		for (int i=1; i<=n; ++i)	{
			for (int j=1; j<=n; ++j)	{
				d[i][j] = (d[i][j] || (d[i][k] && d[k][j]));
			}
		}
	}
}

int main(void)	{
	while (scanf ("%d%d", &n, &m) == 2)	{
		memset (d, false, sizeof (d));
		memset (in, 0, sizeof (in));
		for (int u, v, i=1; i<=m; ++i)	{
			scanf ("%d%d", &u, &v);
			d[u][v] = true;
			in[v]++;
		}
		Floyd_Warshall ();
		int ans = 0;
		for (int i=1; i<=n; ++i)	{
			int c1 = 0, c2 = 0;
			for (int j=1; j<=n; ++j)	{
				if (i == j)	continue;
				if (d[i][j])	c1++;
				if (d[j][i])	c2++;
			}
			if (c1 + c2 == n - 1)	ans++;
		}
		printf ("%d\n", ans);
	}

	return 0;
}