hdu-3342-Legal or Not

最新推荐文章于 2024-08-29 11:53:56 发布

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本文介绍了一个有趣的问题：在一个大型的QQ群中，成员之间形成了一种特殊的“师生”关系，但这种关系可能存在非法的情况，例如形成闭环。文章提供了一种算法来判断这些关系是否合法，并给出了解决方案。

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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

简单的判断是否会成环,若成环输出NO,反之输出YES

注意:输入的可能重复,需要判重

AC:

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
int mp[101][101];
int indegree[101];

using namespace  std;
int n,m;

void topo()
{
    queue<int> que;
    for(int i=0;i<n;i++)
        if(indegree[i]==0)
            que.push(i);
    while(!que.empty())
    {
        int top=que.front();
        que.pop();
        for(int i=0;i<n;i++)
        {
            if(mp[top][i])
            {
                indegree[i]--;
                if(indegree[i]==0)
                    que.push(i);
            }
        }
    }
}

int main()
{
    int a,b,flag;
    while(scanf("%d%d",&n,&m))
    {
        memset(mp,0,sizeof(mp));
        memset(indegree,0,sizeof(indegree));
        flag=0;
        if(n==0&&m==0)
            break;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            if(!mp[b][a])
            {
                mp[b][a]=1;
                indegree[a]++;
            }
        }
        topo();
        for(int i=0;i<n;i++)
        {
           //printf("%d\n",indegree[i]);
           if(indegree[i])
            {
                printf("NO\n");
                flag=1;
                break;
            }
        }
        if(!flag)
            printf("YES\n");
    }
    return 0;
}