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 cowB.

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

Source

题意：有n个牛，编号1~n。现给出m条关系：A B ，说明A比B厉害。现在问有多少个牛能被唯一确定（即这头牛与n-1头牛的关系是唯一确定的）。

解题：一头牛如果能被唯一确定，那么所有的点一定是一个连通块，那么就可能用到并查集 来判断。再接下来就是拓扑排序了。具体看代码。

?

#include
  
   
#include
   
     const int N = 105; bool mapt[N][N],path[N][N]; int n,in[N],father[N]; void init() { for(int i=1;i<=n;i++) { father[i]=i; in[i]=0; for(int j=1;j<=n;j++) mapt[i][j]=path[i][j]=0; path[i][i]=1; } } int findroot(int x) { if(x!=father[x]) father[x]=findroot(father[x]); return father[x]; } void setroot(int x,int y) { x=findroot(x); y=findroot(y); father[x]=y; } int tope() { int a[N],k=0,l=0,ans=0; for(int i=1;i<=n;i++) if(in[i]==0) a[k++]=i; while(l
    
     0) { init(); while(m--) { scanf("%d%d",&a,&b); setroot(a,b); if(mapt[a][b]==0) mapt[a][b]=1,in[b]++; } int k=0; for(int i=1;i<=n;i++) if(father[i]==i) k++; if(k>1)//说明所有的点不是在一个连通块内，所有的点都不能被确定 printf("0\n"); else printf("%d\n",tope()); } } 
    
   
  

?

?