Legal or Not

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 16 Accepted Submission(s) : 2

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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

思路：拓扑排序。若能完成拓扑排序说明此图是有向无环图（DAG），最后所有点的入度都为0，说明至少存在一种拓扑排序，即是关系合理，否则说明是此图有环，明显关系不合理。

 1 #include<stdio.h>
 2 #include<string.h>
 3 int degree[101],vis[101],map[101][101];
 4 int main()
 5 {
 6     int n,m,a,b,i,j,temp,flag;
 7     while(~scanf("%d%d",&n,&m) && n)
 8     {
 9         memset(degree,0,sizeof(degree));
10         memset(vis,0,sizeof(vis));
11         memset(map,0,sizeof(map));
12         flag = 0;
13         while(m--)
14         {
15             scanf("%d%d",&a,&b);
16             if(!map[a][b])　　　　//不计重边；
17             {
18                 degree[b]++;
19                 map[a][b] = 1;
20             }
21         }
22         for(i = 0;i < n;i ++)
23         {
24             for(j = 0;j < n;j ++)  //查找入度为0的点；
25             {
26                 if(degree[j] == 0 && vis[j] == 0) 
27                     temp = j;
28             }
29             vis[temp] = 1; //找到即把此点销毁；
30             for(j = 0;j < n;j ++)
31             {
32                 if(j != temp && map[temp][j] == 1 && vis[j] == 0)   //处理与此点相关的点和边；
33                 {
34                     degree[j]--;
35                     map[temp][j] = 0; //销毁边；
36                 }
37             }
38         }
39         for(i = 0;i < n;i ++)　　　　//验证是否还有入度不为0的点；
40         {
41             if(degree[i])
42             {
43                 flag = 1;
44                 break ;
45             }
46         }
47         if(!flag)
48             printf("YES
");
49         else
50             printf("NO
");
51     }
52     return 0;
53 }