Circle

Circle

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Status

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 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

Hint

 

/*
判环，每个点给出的时候，会调用两次，每个点的祖先都是一个
*/
#include<bits/stdc++.h>
#define N 50
using namespace std;
int bin[N];
int visit[N];
int n,m;
int findx(int x)
{
    while(x!=bin[x])
        x=bin[x];
    return x;
}
void built(int x,int y)
{
    int fx=findx(x);
    int fy=findx(y);
    if(fx!=fy) 
        bin[fx]=fy;
}
bool ok()
{
    for(int i=1;i<=n;i++)
    {
        //cout<<i<<"="<<visit[i]<<endl;
        if(visit[i]!=2)
            return false;
    }    
    for(int i=2;i<=n;i++)
    {
        if(findx(i)!=findx(1))
            return false;
    }
    return true;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(visit,0,sizeof visit);
        for(int i=0;i<n;i++)
            bin[i]=i;
        int x,y;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            visit[x]++;
            visit[y]++;
            built(x,y);
        }
        if(ok())
            puts("YES");
        else
            puts("NO");
    }
}

 

Source

The 10th Zhejiang University Programming Contest