POJ

题意：

确定有多少只牛确定了他们的排名，确定排名必须要进行n-1次比较。

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

5 54 34 23 21 22 5

Sample Output

2

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define MAXV 102#define inf 1<<30using namespace std;int map[MAXV][MAXV];int n,m;int main(){    while(~scanf("%d%d",&n,&m))    {        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++;        }        cout<<ans<<endl;    }    return 0;}