Legal or Not (判断是否存在环)

Legal or Not

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

Total Submission(s) : 61   Accepted Submission(s) : 44

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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

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

Sample Output

YES
NO

Author

QiuQiu@NJFU

Source

HDOJ Monthly Contest – 2010.03.06 

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int indegree[];
int map[][];
int n,m;
void topu()
{
    int p=;
   for(int i=;i<n;i++)
   {
       for(int j=;j<n;j++)
       {
           if(indegree[j]==)
           {

                p++;
               indegree[j]--;
               for(int k=;k<n;k++)
               {
                   if(map[j][k]==)
                   {
                       map[j][k]=;
                      indegree[k]--;
                   }
               }
               break;

           }
       }

   }

   if(p==n)
    printf("YES\n");
   else
    printf("NO\n");
}
int main()
{

    while(~scanf("%d%d",&n,&m))
    {
        memset(map,,sizeof map);
        memset(indegree,,sizeof indegree);
        if(n==&&m==)
            break;
        for(int i=;i<m;i++)
        {
            int q,p;
            scanf("%d%d",&q,&p);
            if(map[q][p]==)
            {
                map[q][p]=;
                indegree[p]++;
            }
        }
      topu();
    }

    return ;
}