ZOJ 3321 Circle【并查集】

最新推荐文章于 2021-08-04 15:15:41 发布

转载最新推荐文章于 2021-08-04 15:15:41 发布 · 118 阅读

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原文链接：http://www.cnblogs.com/wuyuewoniu/p/4254667.html

本文介绍了一个简单的算法，用于判断无向图是否仅由一个环组成。通过输入节点数量及连接情况，利用并查集确定图中各节点是否属于同一连通块且每个节点的度数均为2。

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解题思路：给定n个点，m条边，判断是否构成一个环

注意到构成一个环，所有点的度数为2，即一个点只有两条边与之相连，再有就是判断合并之后这n个点是否在同一个连通块

Circle

Time Limit: 1 Second Memory Limit: 32768 KB

Your task is so easy. I will give you an undirected graph, and you just need to tell me whether the graph is just a circle. A cycle is three or more nodes V₁, V₂, V₃, ... V_k, such that there are edges between V₁ and V₂, V₂ and V₃, ... V_k and V₁, with no other extra edges. The graph will not contain self-loop. Furthermore, there is at most one edge between two nodes.

Input

There are multiple cases (no more than 10).

The first line contains two integers n and m, which indicate the number of nodes and the number of edges (1 < n < 10, 1 <= m < 20).

Following are m lines, each contains two integers x and y (1 <= x, y <= n, x != y), which means there is an edge between node x and node y.

There is a blank line between cases.

Output

If the graph is just a circle, output "YES", otherwise output "NO".

Sample Input

Sample Output

YES
NO

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
using namespace std;
int degree[10010],pre[10010];

int find(int root){ return root == pre[root] ? root : pre[root] = find(pre[root]); }
void unionroot(int x,int y)
{
	int root1=find(x);
	int root2=find(y);
	if(root1!=root2)
	pre[root1]=root2;
}

int main()
{
	int m,n,u,v,i,j;
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		int flag=1;
		memset(degree,0,sizeof(degree));
		for(i=1;i<=10010;i++)
		pre[i]=i;
		for(i=1;i<=m;i++)
		{
			scanf("%d %d",&u,&v);
			degree[u]++;
			degree[v]++;
			unionroot(u,v);	
		}
		
		for(i=1;i<=n;i++)
		{
			if(degree[i]!=2)
			{
				flag=0;
				break;
			}
			if(find(i)!=find(n))
			{
				flag=0;
				break;
			}
		}
		if(flag)
		printf("YES\n");
		else
		printf("NO\n");		
	}
}

转载于:https://www.cnblogs.com/wuyuewoniu/p/4254667.html