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, ... Vk, such that there are edges between V1 and V2, V2 and V3, ... Vk 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\n");
else
printf("NO\n");
}
}