poj 3713(判断无向图是否三连通)

三连通图检测算法

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本文介绍了一种用于检测图是否达到特定安全级别的算法。该算法通过枚举去除每个顶点并检查剩余图中是否存在割点或保持连通性来确保任意两个城市间至少有三条独立路径。适用于评估网络安全系统的可靠性。

Transferring Sylla

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1160		Accepted: 270

Description

After recapturing Sylla, the Company plans to establish a new secure system, a transferring net! The new system is designed as follows:

The Company staff choose N cities around the nation which are connected by "security tunnels" directly or indirectly. Once a week, Sylla is to be transferred to another city through the tunnels. As General ordered, the transferring net must reach a certain security level that there are at least 3 independent paths between any pair of cities a, b. When General says the paths are independent, he means that the paths share only a and b in common.

Given a design of a transferring net, your work is to inspect whether it reaches such security level.

Input

The input consists of several test cases.
For each test case, the first line contains two integers, N ≤ 500 and M ≤ 20000. indicating the number of cities and tunnels.
The following M lines each contains two integers a and b (0 ≤ a, b < N), indicating the city a and city b are connected directly by a tunnel.

The input ends by two zeroes.

Output

For each test case output "YES" if it reaches such security level, "NO" otherwise.

Sample Input

Sample Output

YES
NO
NO

Source

POJ Founder Monthly Contest – 2008.12.28, Dagger

题目： http://poj.org/problem?id=3713

分析：三连通没想法阿，看别人的题解都是枚举去掉一个点，然后判断是否存在割点或并且是否连通，如果存在割点或者不连通，就不是三连通。。。很奇怪阿，时间太恶心了。不知道那几个几百ms的怎么来的

代码：

#include<cstdio>
using namespace std;
const int mm=44444;
const int mn=555;
int t[mm],p[mm];
int h[mn],dfn[mn],low[mn],du[mn];
int i,j,k,n,m,idx;
bool dfs(int u,int fa)
{
    dfn[u]=low[u]=++idx;
    for(int i=h[u],v,son=0;i>=0;i=p[i])
        if(!dfn[v=t[i]])
        {
            ++son;
            if(dfs(v,u))return 1;
            if(fa==-1&&son>1||fa!=-1&&dfn[u]<=low[v])return 1;
            if(low[u]>low[v])low[u]=low[v];
        }
        else if(v!=fa&&low[u]>dfn[v])low[u]=dfn[v];
    return 0;
}
bool tarjan()
{
    int i,j,k;
    for(i=0;i<n;++i)
        if(du[i]<3)return 0;
    for(i=0;i<n;++i)
    {
        for(j=idx=0;j<n;++j)dfn[j]=0;
        dfn[i]=n+n;
        for(j=k=0;j<n;++j)
            if(!dfn[j])
            {
                if(++k>1)return 0;
                if(dfs(j,-1))return 0;
            }
    }
    return 1;
}
int main()
{
    while(scanf("%d%d",&n,&m),n+m)
    {
        for(i=k=0;i<n;++i)h[i]=-1,du[i]=0;
        while(m--)
        {
            scanf("%d%d",&i,&j);
            t[k]=j,p[k]=h[i],h[i]=k++;
            t[k]=i,p[k]=h[j],h[j]=k++;
            ++du[i],++du[j];
        }
        puts(tarjan()?"YES":"NO");
    }
    return 0;
}