POJ 3660 Cow Contest

POJ 3660 Cow Contest

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

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

// 牛 A能打敗 牛B 問有多少個牛的順序能被確定。

//記錄每個牛的度，如果等於N-1 表示能確定順序。

#include 
#include 
#include 
#include 
#include 
#include 
#define N 109
using namespace std;

int mp[N][N];
int n,m;
int a,b;
int deg[N];


int main()
{

    while(~scanf("%d %d",&n,&m))
    {
        memset(deg,0,sizeof deg);
        memset(mp,0,sizeof mp);

        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&a,&b);
            mp[a][b]=1;
        }

        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        for(int k=1;k<=n;k++)
        if(mp[j][i] && mp[i][k])
        mp[j][k]=1;

        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(mp[i][j])
            {
                deg[i]++;
                deg[j]++;
            }
        }
        int ans=0;

        for(int i=1;i<=n;i++)
        if(deg[i]==n-1)
        ans++;

        printf("%d\n",ans);
    }

    return 0;

}