题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3342

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7416    Accepted Submission(s):
3541

Problem DescriptionACM-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.

 InputThe 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. OutputFor each test case, print in one line the judgement of
the messy relationship.
If it is legal, output “YES”, otherwise “NO”. Sample Input3 20 11 22 20 11 00 0 Sample OutputYESNO 题目大意：有N个人，给出M组两个人之间的师徒关系，问是否存在悖论，即某个人的徒弟又是自己的师傅。 解题思路：用拓扑排序判断图中是否存在环即可 AC代码：

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 using namespace std;
 queue <int > q;
 int line[][];
 int in[];
 int main ()
 {
     int i,j,m,n,num,u,v;
     while (scanf("%d%d",&n,&m)!=EOF)
     {
         if (n == )
             break;
         memset(line,,sizeof(line));
         memset(in,,sizeof(in));         for (i = ; i < m; i ++)
         {
             scanf("%d%d",&u,&v);
             if (line[u][v]==)  //判断重边
             {
                 line[u][v] ++;
                 in[v] ++;
             }
         }
         for (i = ; i < n; i ++)
             if (in[i] == ) q.push(i);  //入度为0的存入队列
         num = ;
         while (!q.empty())
         {
             int k = q.front();
             q.pop();
             num ++;
             for (i = ; i < n; i ++)
             {
                 if (line[k][i] > )  //删掉与该入度为0的节点相连的边
                 {
                     in[i] --;
                     if (in[i] == )
                         q.push(i);
                 }
             }
         }
         if (num == n)    //判断是否存在环
             printf("YES\n");
         else
             printf("NO\n");
     }
     return ;
 }