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  • 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, ... 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
    

    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
     
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  • 原文地址:https://www.cnblogs.com/wuwangchuxin0924/p/5936259.html
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