HDOJ 5222 Exploration 并查集+拓扑排序 找环

Exploration

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 715    Accepted Submission(s): 197

Problem Description

Miceren likes exploration and he found a huge labyrinth underground! 

This labyrinth has  N  caves and some tunnels connecting some pairs of caves. 

There are two types of tunnel, one type of them can be passed in only one direction and the other can be passed in two directions. Tunnels will collapse immediately after Miceren passing them. 

Now, Miceren wants to choose a cave as his start point and visit at least one other cave, finally get back to start point. 

As his friend, you must help him to determine whether a start point satisfing his request exists.

 

Input

The first line contains a single integer  T , indicating the number of test cases.

Each test case begins with three integers  N, M1, M2 , indicating the number of caves, the number of undirectional tunnels, the number of directional tunnels. 

The next  M1  lines contain the details of the undirectional tunnels. Each line contains two integers  u, v  meaning that there is a undirectional tunnel between  u, v . ( u ≠ v ) 

The next  M2  lines contain the details of the directional tunnels. Each line contains integers  u, v  meaning that there is a directional tunnel from  u  to  v . ( u ≠ v )

T  is about 100.

1 ≤ N,M1,M2 ≤ 1000000.

There may be some tunnels connect the same pair of caves.

The ratio of test cases with  N > 1000  is less than 5%.

 

Output

For each test queries, print the answer. If Miceren can do that, output "YES", otherwise "NO".

 

Sample Input

   
   
   
   

    
    
    
    2
5 2 1
1 2
1 2
4 5
4 2 2
1 2
2 3
4 3
4 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    YES
NO


    
    
    
    
 
     
 
      Hint 
     
If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++. 
    
    
    
    
     
   
   
   
   

 

Source

赛码"BestCoder"杯中国大学生程序设计冠军赛

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年05月07日 星期四 22时48分45秒
File Name     :HDOJ5222.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

#pragma comment(linker, "/STACK:102400000,102400000") 

using namespace std;

const int maxn=1001000;

int fa[maxn];

int find(int x)
{
	if(fa[x]==x) return x;
	return fa[x]=find(fa[x]);
}

bool Union(int a,int b)
{
	a=find(a); b=find(b);
	if(a!=b) 
	{
		fa[a]=b; return true;
	}
	return false;
}

int n,m1,m2;

struct Edge
{
	int to,next;
}edge[2*maxn];

int Adj[maxn],Size;
int degree[maxn];
bool vis[maxn];

void init()
{
	for(int i=0;i<=n+10;i++) fa[i]=i;
	memset(Adj,-1,sizeof(Adj)); Size=0;
	memset(degree,0,sizeof(degree));
	memset(vis,false,sizeof(vis));
}

void Add_Edge(int u,int v)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	Adj[u]=Size++;
}


int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    
	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d%d",&n,&m1,&m2);
		init();
		bool flag=false;
		for(int i=0;i<m1;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			if(Union(u,v)==false) flag=true;
		}
		
		for(int i=0;i<m2;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			if(flag) continue;
			u=find(u); v=find(v);
			if(u!=v) 
			{
				/// u--->v
				Add_Edge(v,u);
				degree[u]++;
			}
			else flag=true;
		}

		if(flag)
		{
			puts("YES"); continue;
		}

		queue<int> q;

		for(int i=1;i<=n;i++)
		{
			int u=find(i);
			if(vis[u]) continue;
			if(degree[u]==0)
			{
				q.push(u);
				vis[u]=true;
			}
		}

		while(!q.empty())
		{
			int u=q.front(); q.pop();
			u=find(u);
			for(int i=Adj[u];~i;i=edge[i].next)
			{
				int v=edge[i].to;
				v=find(v);
				degree[v]--;
				if(degree[v]==0) 
				{
					q.push(v);
					vis[v]=true;
				}
			}
		}

		bool ck=true;
		for(int i=1;i<=n&&ck;i++)
		{
			if(vis[find(i)]==false) ck=false;
		}

		if(ck) puts("NO");
		else puts("YES");
	}

    return 0;
}