hiho215 Circle Detect

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

You are given a directed graph G which has N nodes and M directed edges. Your task is to detect whether it contains any circle.

输入

The first line contains an integer T denoting the number of test cases. (1 <= T <= 5)

For each test case the first line contains two integers N and M. (1 <= N, M <= 100000)

Then follows M lines. Each contains two integers u and v denoting there is an edge from u to v. (1 <= u, v <= N)

输出

For each test case output "YES" or "NO" denoting whether there is a circle in the graph.

样例输入

样例输出

YES  
NO

题意分析:

1.题是什么?

这道题很直白,就是给你一个有向图,让你判断是否存在环.

2.思路

最开始想成并查集去了,后来发现是有向图.....如果是无向图可以已并查集快速搞定,有向图不行,下面举出特例:

对于这个最简单的有向图,明显不该被认定为环,然而如果这四条边是无向边,自然这是个环.

然后是思路,对每个结点做dfs,dfs的同时做标记,如果发现dfs到了刚才走过的结点,自然就成环了,关键在于dfs回溯时要对标记做一个清除,这样才能正确.

ac代码:

#include <iostream> 
#include <stdio.h>
#include <vector>
using namespace std;

const int maxn=100005;
vector<int> children[maxn]; //children[i]存储i的所有子节点 
int flag[maxn]; //flag[i]为0表示未被dfs过,为1表示刚刚经过过它,为2表示是已经dfs完所有子节点无环的安全结点

bool dfs(int pos){
	flag[pos]=1;
	int num=children[pos].size();
	for(int i=0;i<num;i++){
		int t=children[pos][i];
		if(flag[t]==1) return true;
		if(flag[t]==0&&dfs(t)) return true;
	}
	flag[pos]=2; 
	return false;
} 

void solve(){
	int t,n,m,t1,t2;
	bool existcircle;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&m);
		
		//初始化 
		for(int i=1;i<=n;i++){
			children[i].clear();
			flag[i]=0;	
		}
		existcircle=false;
		
		//读入 
		for(int i=0;i<m;i++){
			scanf("%d%d",&t1,&t2);
			children[t1].push_back(t2);
		}
		
		//dfs 
		for(int i=1;i<=n;i++){
			if(flag[i]==0){
				if(dfs(i)){
					existcircle=true;
					break;
				}
			}
		}
		
		//输出 
		if(existcircle) printf("YES\n");
		else printf("NO\n");
	}
}

int main(){
	solve();
	
	return 0;
}