Legal or Not 【topo 判定是否有环】

最新推荐文章于 2019-07-20 09:54:08 发布

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最新推荐文章于 2019-07-20 09:54:08 发布

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分类专栏：拓扑排序

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本文介绍了一个基于图论的问题：在一个有向图中判断‘师傅’与‘徒弟’之间的关系是否合法，即是否存在非法的互为师傅与徒弟的关系。通过使用拓扑排序的方法来解决该问题，确保了每一对‘师傅’与‘徒弟’的关系符合合法性要求。

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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
3 2
0 1
1 2
2 2
0 1
1 0
0 0
Sample Output
YES
NO
给定一个有向图，问其中是不是有环，
裸topo
看代码

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MAXN =2000+100;
const int MAXM=1000000;
vector<int>G[MAXN];
int n,m;
int in[MAXN];
void init(){
    for(int i=0;i<=n;i++)
    {
        G[i].clear();
        in[i]=0;
    }
}
void getmap(){
    int a,b;
    while(m--){
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        in[b]++;
    }
}
bool topo(){
    queue<int>Q;
    for(int i=0;i<n;i++)
        if(in[i]==0) Q.push(i);
        int num=0;
    while(!Q.empty()){
        int now=Q.front();Q.pop();num++;
        for(int i=0;i<G[now].size();i++){
            if(--in[G[now][i]]==0)
                Q.push(G[now][i]);
        }
    }
    return num!=n;
}
int main(){
     while(scanf("%d%d",&n,&m)&&(n||m)){
        init();
        getmap();
        printf("%s\n",!topo()?"YES":"NO");
     }
}