Leagal or Not —— 拓扑排序（王道）

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

拓扑排序，题意是给出多个人之见的关系，让你判断是否有环。

#include <iostream>
#include<queue>
#include<cstdio>
using namespace std;
vector<int> edge[];//邻接链表，只需保存与其邻接的节点编号
queue<int> Q;//保存入度为0的节点的队列
int main()
{
    int inDegree[];//统计每个节点的入度
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF){
        if(n== || m==)
            break;
        for(int i=;i<n;i++){//初始化
            inDegree[i]=;
            edge[i].clear();
        }
        while(m--){
            int a,b;
            scanf("%d %d",&a,&b);
            inDegree[b]++;//b的入度+1
            edge[a].push_back(b);//a的邻接表中添加b
        }
        while(!Q.empty())//清空队列
            Q.pop();
        for(int i=;i<n;i++){
            if(inDegree[i]==)//入度为0的放入队列
                Q.push(i);
        }
        int cnt=;
        while(!Q.empty()){
            int nowP=Q.front();
            Q.pop();
            cnt++;//被确定的节点个数加一
            for(int i=;i<edge[nowP].size();i++){//将该节点以及以其为弧尾的所有边去除
                inDegree[edge[nowP][i]]--;//该边的入度减一
                if(inDegree[edge[nowP][i]]==)//该点的入度变为0
                    Q.push(edge[nowP][i]);//把这个点放入队列当中
            }
        }
        puts(cnt==n?"YES":"NO");//所有节点都被确定拓扑序列，原图为有向无环图；否则非有向无环图
    }
    return ;
}