Harry and Magical Computer

In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

Input

There are several test cases, you should process to the end of file. 
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤100001≤n≤100,1≤m≤10000 
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n1≤a,b≤n

Output

Output one line for each test case. 
If the computer can finish all the process print "YES" (Without quotes). 
Else print "NO" (Without quotes).

Sample Input

3 2
3 1
2 1
3 3
3 2
2 1
1 3

Sample Output

YES
NO

测试样例分析：

 

#include <stdio.h>
#include <string.h> 
#define MAX 100+1 
const int INF = 0x3f3f3f3f;

bool G[MAX][MAX];//用于判断有无连线 
int in[MAX];//in[i] 表示i点的入度 
 
bool topoSort(int n){//n个点 
	int cnt = 0;//作为指针作用 
	int k;
	for(int t = 1;t<=n;t++){//n门课，需要找n遍 
		k = -1;
		for(int i = 1;i<=n;i++){
			if(in[i]==0){
				k = i;
				break;
			}
		}
		//说明一趟下来没有找到入度为0的点，即存在环 
		if(k == -1) return false;
		
		for(int i = 1;i<=n;i++){
			if(G[k][i]){ //k --> i 即i为k课程的后续点 
				in[i]--;//那么直接后续点的入度-1 
			}
		}
		in[k] = -1;//用-1表示这个点被选过了 
	}
	return true; 
}
 
int main(){
	int n,m;//n个点，m对关系 
	while(scanf("%d%d",&n,&m)!=EOF){
		//初始化 
		memset(in,0,sizeof(in));
		memset(G,false,sizeof(G));
		
		int u,v;//v-->u 即v完成才能到u 
		for(int i = 0;i<m;i++){
			scanf("%d%d",&u,&v);
			if(!G[v][u]){
				//v-->u即为u的入度+1，v的出度+1 
				G[v][u] = true;
				in[u]++;
			}
		}
		
		if(topoSort(n)){
			printf("YES\n");
		}
		else{
			printf("NO\n");
		}
	}
	return 0;
}