Legal or Not

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5942 Accepted Submission(s): 2752

Problem Description

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.

Input

The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.

Output

For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".

Sample Input

Sample Output

YES
NO

题意：输入数据n，m，表示有n个人接下来m行，每行输入x，y表示x是y的师父。
假设A是B的师父B是C的师父，则A是C的师父
假设A是B的师父，B又是A的师父则不合法输出No，假设合法输出YES

本题的关键是怎么推断成环，我的方法是通过拓扑排序的方法把输入的数据排序，由拓扑排序可知。每次都能找到一个数放到该序列里，假设有一步找不到，就说明成环了。输出NO。假设结束时还没出现上面情况，输出YES。

#include<stdio.h>  
#include<string.h>  
int degree[1000];  
int side[1000];  
int map[510][501],list,line;  
void TO()  
{  
    int i,j=0,k=100000,t;  
    for(i=1;i<=line;i++)  
    {  
        for(t=1;t<=line;t++)  
        if(degree[t]==0)//找没有前驱的点   
        {  
            k=t;  
            break;  
        } 
        else k=100000;
        if(k==100000)//推断是否成环
        {
        	printf("NO
");
        	return ;
        }
        side[j++]=k;//记录该点   
        degree[k]=-1;//把该点去掉   
        for(int v=1;v<=line;v++)  
        if(map[k][v]) degree[v]--;//前驱与后面的连线消失   
    }  
  //  if(j==line)
   printf("YES
");
    //else printf("NO
");
}  
int main()  
{  
    while(scanf("%d%d",&line,&list)!=EOF)  
    {  
       if(line==0&&list==0)
       break;
        int j,a,b;  
        memset(map,0,sizeof(map));  
        memset(degree,0,sizeof(degree));  
        for(j=0;j<list;j++)  
        {  
            scanf("%d%d",&a,&b); 
			a=a+1;b=b+1; 
            if(map[a][b]==0)//防止出现反复如1,2；1,2这种情况   
            {  
                map[a][b]=1;// 与前一步的关系；   
                degree[b]++;//前驱的数量   
            }  
        }  
        TO();  
    }  
    return 0;  
}