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：
在已知部分大小关系的基础上，推断有几个排名确定。
Ps：
当一个点与其他所有点的关系确定，他的排名就确定。用Floyd即可算出。
代码：
#include<stdio.h> #include<iostream> #include<string.h> using namespace std; const int N=110; int mmp[N][N]; int main() { int n,m; while(~scanf("%d%d",&n,&m)) { int u,v; memset(mmp,0,sizeof(mmp)); for(int i=1; i<=m; i++) { scanf("%d%d",&u,&v); mmp[u][v]=1; } for(int k=1; k<=n; k++) for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) if(mmp[i][k]&&mmp[k][j]) mmp[i][j]=1; int ans=0,j; for(int i=1; i<=n; i++) { for(j=1; j<=n; j++) { if(i==j) continue; if(mmp[i][j]==0&&mmp[j][i]==0) break; } if(j>n) ans++; } printf("%d\n",ans); } }