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
floyd拓扑判断连通性，如果出度+入度=n-1则可判断排名。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <vector> 6 #include <queue> 7 #include <set> 8 #include <map> 9 #include <string> 10 #include <cmath> 11 #include <stdlib.h> 12 #define N 105 13 using namespace std; 14 int n,m; 15 int g[N][N]; 16 int u,v; 17 void floyd() 18 { 19 for(int k = 1;k<=n;k++) 20 for(int i = 1;i<=n;i++) 21 for(int j = 1;j<=n;j++) 22 g[i][j] |= g[i][k]&&g[k][j]; 23 } 24 int main() 25 { 26 //freopen("caicai.txt","r",stdin); 27 scanf("%d%d",&n,&m); 28 for(int i = 0;i<m;i++) 29 { 30 scanf("%d%d",&u,&v); 31 g[u][v] = 1; 32 } 33 floyd(); 34 int ans = 0; 35 for(int i = 1;i<=n;i++) 36 { 37 int cnt = 0; 38 for(int j = 1;j<=n;j++) 39 { 40 if(i==j) continue; 41 if(g[i][j]||g[j][i]) 42 cnt++; 43 } 44 if(cnt==n-1) 45 ans++; 46 } 47 cout<<ans<<endl; 48 return 0; 49 }