POJ-3660-Cow Contest

给出牛之间的强弱关系，让你确定有多少头牛能够确定其排名。

用Floyd做，对每给的一个胜负关系连一条边，最后跑一次Floyd，然后判断一头牛所确定的关系是否是n-1次，若是，则这头牛的排名可以确定

有n只奶牛，有n个连续的实力，如果u的实力大于v的实力，就能打赢它，*        然后给定m种关系，求最后能确定其排名的奶牛个数。

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 cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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

5 54 34 23 21 22 5

Sample Output

2
#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define maxn=102;int map[102][102];int n,m;int main(){while(scanf("%d%d",&n,&m)!=EOF){memset(map,0,sizeof(map));for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);map[a][b]=1;}for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(map[i][k]&&map[k][j])map[i][j]=1;int ans=0;for(int i=1;i<=n;i++){int res=n-1;for(int j=1;j<=n;j++)if(map[i][j]||map[j][i])res--;if(!res)ans++;}printf("%d\n",ans);}return 0;}