ZOJ 3321 Circle



I - Circle

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld
& %llu

Submit Status Practice ZOJ
3321

Description

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 nodey.

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

给你一幅无向图，让你判断是不是圆。。

成圆条件有两个。

第一：每一个点的入度只能为2.

第二：所有点属于一个集合。

上代码；

#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#include <iostream>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <time.h>
#include <stdlib.h>
int p[26];
int v[26];
int find(int x)
{
    if(p[x]!=x)
        p[x]=find(p[x]);
    return p[x];
}
int hebing(int x,int y)
{
    return p[x]=y;
}
int main()
{
    int n,m,i,j,a,b;
    while(cin>>n>>m)
    {
        memset(v,0,sizeof(v));
        for(i=1; i<=25; i++)
            p[i]=i;
        while(m--)
        {
            cin>>a>>b;
            v[a]++;  //入度
            v[b]++;
            a=find(a);
            b=find(b);
            if(a!=b)
                hebing(a,b);   //合并、
        }
        for(i=1; i<=n; i++)
        {
            if(p[1]!=p[i]|| v[i]!=2)
                break;
        }
       // cout<<i<<endl;
        if(i==n+1)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}