Legal or Not

描述：

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.

输入：

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.

输出：

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

样例输入：

3 2
0 1
1 2
2 2
0 1
1 0
0 0

样例输出：

YES
NO

题目大意：

一个师傅可以有很多的徒弟，一个徒弟可以有很多师傅这些关系都是合理的，但不可以a是b的师傅而b又是a的师傅。试判断对于给出的数据关系是否合理。

代码如下：

//拓扑排序：由AOV网构造拓扑序列的拓扑排序算法主要是循环执行以下两步，直到不存在入度为0的顶点为止。//(1) 选择一个入度为0的顶点并输出之；//(2) 从网中删除此顶点及所有出边。//循环结束后，若输出的顶点数小于网中的顶点数，则输出“有回路”信息，否则输出的顶点序列就是一种拓扑序列 #include<stdio.h>#include<string.h>int qwe[105][105];int mask[105];  int sove(int a)                      //拓扑排序 {    int mak;    for(int i=0;i<a;i++)    {        mak=a;        for(int j=0;j<a;j++)        {            if(mask[j]==0)   //先找到度为0的点             {                mask[j]--;                mak=j;                break;            }        }        if(mak==a) return 0;        for(int j=0;j<a;j++)        {            if(qwe[mak][j]>0)                {                mask[j]--;            }        }    }    return 1;}int main(){    int a,b,c,d;    while(scanf("%d %d",&a,&b),a!=0||b!=0)    {        memset(qwe,0,sizeof(qwe));             //初始化         memset(mask,0,sizeof(mask));        for(int i=0;i<b;i++)        {            scanf("%d %d",&c,&d);            if(qwe[c][d]==0)            {                qwe[c][d]=1;//置为1表示c是d的师傅                 mask[d]++;         //d的入度加一             }        }        int flag=sove(a);        if(flag) printf("YES\n");        else printf("NO\n");    }    return 0;}