| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12162 | Accepted: 6756 |
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先用Floyd算法找出人人间的关系，然后如果关系数等于n-1，那么这个人的排名就能确定
AC代码：
#include<cstdio>
int e[110][110];
int main(){int n,m;scanf("%d%d",&n,&m);while(m--){int a,b;scanf("%d%d",&a,&b);e[a][b]=1;}for(int k=1;k<=n;k++){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(e[i][k]==1&&e[k][j]==1){e[i][j]=1;}}}}int sum=0;for(int i=1;i<=n;i++){int t=0;for(int j=1;j<=n;j++){t+=e[i][j]+e[j][i];}if(n-1==t){sum++;}}printf("%d",sum);return 0;
}