杭电3342Legal or Not

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5982    Accepted Submission(s): 2771

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

3 20 11 22 20 11 00 0

 

Sample Output

YESNO

 

判断有向图里边有没有环，有人说用并查集为什么不能确定环，因为并查集是无向图哦，比如说并查集里边1,2  2,3   1,3  是环，但在有向图里边不是环。

附ac代码：

#include<stdio.h>#include<string.h>int a[110][110],topo[110];int main(){int m,n,i,j,k,x,y,flag;while(scanf("%d%d",&m,&n),m){memset(a,0,sizeof(a));memset(topo,0,sizeof(topo));for(i=0;i<n;i++){scanf("%d%d",&x,&y);if(!a[x][y]){a[x][y]=1;topo[y]++;}}flag=0;for(i=0;i<m;i++) {if(!topo[i]){topo[i]=1000;//只要不等于0，等于几都行 flag++;//也就是说，一直判断着，如果topo都等于0，就判断成功，flag一定等于m， for(j=0;j<m;j++)if(a[i][j])topo[j]--;i=-1;//如果存在 1是2的 老师，2又是1的老师这种情况，到最后topo【1】==1，topo[2]==1,解不开环 }//就判断失败，i继续加而flag不加，flag一定加不到m }if(flag==m)printf("YES\n");elseprintf("NO\n");}return 0;}