UVA-11396 Claw Decomposition (二分图判定)

题目大意:给一张无向图,能否把它分成若干个“爪”,即,一个点有三个子节点。

题目分析:每个点的度数3是已知的,只需判断一下是不是二分图即可。

代码如下:

# include<iostream>
# include<cstdio>
# include<vector>
# include<cstring>
# include<algorithm>
using namespace std;
# define REP(i,s,n) for(int i=s;i<n;++i)
# define CL(a,b) memset(a,b,sizeof(a))

const int N=305;
vector<int>G[N];
int vis[N];

bool dfs(int u)
{
    for(int i=0;i<G[u].size();++i){
        int v=G[u][i];
        if(v==u) continue;
        if(vis[u]==vis[v]) return false;
        if(vis[v]) continue;
        vis[v]=3-vis[u];
        if(!dfs(v)) return false;
    }
    return true;
}

bool judge(int n)
{
    CL(vis,0);
    REP(i,0,n) if(!vis[i]){
        ++vis[i];
        if(!dfs(i)) return false;
    }
    return true;
}

int main()
{
    int n,a,b;
    while(scanf("%d",&n)&&n)
    {
        REP(i,0,n) G[i].clear();
        while(scanf("%d%d",&a,&b)&&(a+b))
        {
            --a,--b;
            G[a].push_back(b);
            G[b].push_back(a);
        }
        if(judge(n))
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/20143605--pcx/p/4923246.html