hdu 3342 Legal or Not

最新推荐文章于 2021-05-06 20:32:40 发布

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最新推荐文章于 2021-05-06 20:32:40 发布

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分类专栏：拓扑排序

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本文介绍了一个算法问题，即如何通过图论中的拓扑排序来判断一组师生关系是否合法。问题来源于一个大型QQ群ACM-DIY，群内的成员互相帮助，并形成了复杂的师生关系网。文章提供了一段C++代码实现，用于检测输入的师生关系是否存在非法的循环关系。

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Legal or Not

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2602 Accepted Submission(s): 1158

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

Sample Output

YES
NO

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxx=105;
//degree[]	每个结点的入度
//f[]	每个结点所在的层
int degree[maxx];
int f[maxx];
int map[maxx][maxx];
int top;
int n,m;
bool Toplogical_sort()
{
    int i,j;
    bool p=true;
    top=0;
    int tmp=0;
    while(p)
    {

        p=false;
        top++;
        tmp=-1;
        for(i=0; i<=n-1; i++)
            if(degree[i]==0)
            {
                p=true;
                f[i]=top;
                tmp=i;
                break;
            }
        if(tmp==-1)
            continue;
        for(j=0; j<n; j++)
        {
            if(map[tmp][j])
                degree[j]--;
            degree[tmp]=-1;
        }
    }
    top--;
    if(top<n)
        return false;
    return true;
}
int main()
{
    while(scanf("%d%d",&n,&m))
    {
        if(n==0)
        break;
        memset(map,0,sizeof(map));
        memset(f,0,sizeof(f));
        memset(degree,0,sizeof(degree));
        int a,b;
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&a,&b);
            if(!map[a][b])
            {
                map[a][b]=1;
                degree[b]++;
            }
        }
        if(Toplogical_sort())
        {
            printf("YES\n");
        }
        else printf("NO\n");
    }
    return 0;
}