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++
1 #include<iostream> 2 #include<algorithm> 3 #include<vector> 4 #include<queue> 5 6 using namespace std; 7 const int INF = 0x3f3f3f3f; 8 int dis[110][110]; 9 int n, m; 10 void floyed() 11 { 12 for (int k = 1; k <= n; k++) 13 for (int i = 1; i <= n; i++) 14 for (int j = 1; j <= n; j++) { 15 if ((dis[i][k] == 1 && dis[k][j] == 1) || (dis[i][k] == 2 && dis[k][j] == 2)) 16 dis[i][j] = dis[i][k]; //判断ij之间的关系 17 } 18 } 19 int main() 20 { 21 ios::sync_with_stdio(false); 22 while (cin >> n >> m) { 23 memset(dis, 0, sizeof(dis)); 24 for (int a, b, i = 1; i <= m; i++) { 25 cin >> a >> b; 26 dis[a][b] = 1;//a胜b 27 dis[b][a] = 2;//a输给b 28 } 29 floyed(); 30 int ans = 0; 31 for (int i = 1; i <= n; i++) { 32 int cnt = 0; 33 for (int j = 1; j <= n; j++) { 34 if (i == j)continue; 35 if (dis[i][j])cnt++; 36 } 37 if (cnt == (n - 1))ans++; 38 } 39 cout << ans << endl; 40 } 41 return 0; 42 }