Cow Contest(传递闭包)

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10450		Accepted: 5841

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

//题意是 n 头牛 m 个牛的实力信息，给出 m 头牛的实力信息后，问多少的牛可以确定排名

//离散里面的传递闭包问题

//三重循环暴力即可

 #include <iostream>
 #include <cstdio>
 using namespace std;
 int f[][],n,m;

 int main()
 {
     cin>>n>>m;
     for (int i=;i<=m;i++)
     {
         int x,y;
         cin >> x >> y;
         f[x][y]=;
     }
     for (int k=;k<=n;k++)
       for (int i=;i<=n;i++)
         for (int j=;j<=n;j++)
           if (f[i][k]+f[k][j]==)
             f[i][j]=;
     int ans=;
     for (int i=;i<=n;i++)
     {
         int total=;
         for (int j=;j<=n;j++)
           if (f[i][j] || f[j][i]) total++;
         if (total==n-) ans++;
     }
     cout << ans << endl;
 }