Legal or Not(HDU 3342)

题目链接

题目描述

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.

输入格式

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.

输出格式

For each test case, print in one line the judgement of the messy relationship.
If it is legal, output “YES”, otherwise “NO”.

输入样例

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

输出样例

YES
NO

分析

题目大意给定n个点，m条边的有向图，判断是否能拓扑排序。
直接套用模板判断即可，若无法进行拓扑排序就代表有环。

源程序

#include <bits/stdc++.h>
#define MAXN 105
using namespace std;
struct Edge{
	int v,next;
	Edge(){};
	Edge(int _v,int _next){
		v=_v,next=_next;
	}
}edge[MAXN];
int EdgeCount,head[MAXN];
int n,m,in[MAXN];
void addEdge(int u,int v)
{
	edge[++EdgeCount]=Edge(v,head[u]);
	head[u]=EdgeCount; 
}
bool TopSort()
{
	queue<int> q;
	for(int i=0;i<n;i++)
		if(!in[i])q.push(i);
	int cnt=0;
	while(!q.empty()){
		int u=q.front();
		q.pop();
		cnt++;
		for(int i=head[u];i;i=edge[i].next){
			int v=edge[i].v;
			in[v]--;
			if(!in[v])q.push(v);
		}
	}
	if(cnt!=n)return false;
	return true;
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF&&n){
		EdgeCount=0;
		for(int i=0;i<n;i++)head[i]=in[i]=0;	//初始化 
		for(int i=1;i<=m;i++){
			int u,v;
			scanf("%d%d",&u,&v);
			in[v]++;
			addEdge(u,v);
		}
		if(TopSort())printf("YES\n");
		else printf("NO\n");
	}	
}