poj 3660(Floyd求传递闭包)

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9317		Accepted: 5249

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

Sample Output

题意：有n头牛比赛，m种比赛结果，最后问你一共有多少头牛的排名被确定了，其中如果a战胜b，b战胜c，则也可以说a战胜c，即可以传递胜负。求能确定排名的牛的数目。

解题思路：如果一头牛被x头牛打败，打败y头牛，且x+y=n-1，则我们容易知道这头牛的排名就被确定了，所以我们只要将任何两头牛的胜负关系确定了，在遍历所有牛判断一下是否满足x+y=n-1，将满足这个条件的牛数目加起来就是所求解。

可以利用传递闭包，将某头牛到其它牛的"可达性"求出。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;const int maxn = 105;
int n,m,map[maxn][maxn];void floyd()
{for(int k = 1; k <= n; k++)for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)map[i][j] = map[i][j] || (map[i][k] && map[k][j]);
}int main()
{int u,v;while(scanf("%d%d",&n,&m)!=EOF){memset(map,0,sizeof(map));for(int i = 1; i <= m; i++){scanf("%d%d",&u,&v);map[u][v] = 1;}floyd();int ans = 0,tmp;for(int i = 1; i <= n; i++){tmp = 0;for(int j = 1; j <= n; j++)if(i != j && (map[i][j] || map[j][i]))tmp++;if(tmp == n-1) ans++;}printf("%d\n",ans);}return 0;
}

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