POJ 3660 Cow Contest（floyed运用）

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

题目大意：有n头奶牛，奶牛之间有m个关系，每次输入的x，y代表x胜过y，求出能够确定当前奶牛和其他所有奶牛的关系的奶牛有几头。
思路：对于每两个奶牛之间有三种关系， 1.没关系 2. a胜过b 3. a输给b，我们用dis[i][j] 代表第i只奶牛和第j只奶牛的关系。我们首先可以对开始输入的奶牛的关系建图，之后用floyed跑一遍图，遍历完所有点点之间的关系，最后判断每一个点，若与其他n-1个点都有关系则ans++

 #include<iostream>
 #include<algorithm>
 #include<vector>
 #include<queue>

 using namespace std;
 const int INF = 0x3f3f3f3f;
 int dis[][];
 int n, m;
 void floyed()
 {
     for (int k = ; k <= n; k++)
         for (int i = ; i <= n; i++)
             for (int j = ; j <= n; j++) {
                 if ((dis[i][k] ==  && dis[k][j] == ) || (dis[i][k] ==  && dis[k][j] == ))
                     dis[i][j] = dis[i][k]; //判断ij之间的关系
             }
 }
 int main()
 {
     ios::sync_with_stdio(false);
     while (cin >> n >> m) {
         memset(dis, , sizeof(dis));
         for (int a, b, i = ; i <= m; i++) {
             cin >> a >> b;
             dis[a][b] = ;//a胜b
             dis[b][a] = ;//a输给b
         }
         floyed();
         int ans = ;
         for (int i = ; i <= n; i++) {
             int cnt = ;
             for (int j = ; j <= n; j++) {
                 if (i == j)continue;
                 if (dis[i][j])cnt++;
             }
             if (cnt == (n - ))ans++;
         }
         cout << ans << endl;
     }
     return ;
 }