ZOJ 3321 Circle【并查集】

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

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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

3 3
1 2
2 3
1 3

4 4
1 2
2 3
3 1
1 4

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
");
		else
		printf("NO
");		
	}
}