hdu 4587 TWO NODES（tarjan算法 删两个点求最多剩余联通分量）

TWO NODES

Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3091    Accepted Submission(s): 974

 

Problem Description

Suppose that G is an undirected graph, and the value of stab is defined as follows:

Among the expression,G-i, -j is the remainder after removing node i, node j and all edges that are directly relevant to the previous two nodes. cntCompent is the number of connected components of X independently.
Thus, given a certain undirected graph G, you are supposed to calculating the value of stab.

 

Input

The input will contain the description of several graphs. For each graph, the description consist of an integer N for the number of nodes, an integer M for the number of edges, and M pairs of integers for edges (3<=N,M<=5000).
Please note that the endpoints of edge is marked in the range of [0,N-1], and input cases ends with EOF.

 

Output

For each graph in the input, you should output the value of stab.

 

Sample Input

4 5 0 1 1 2 2 3 3 0 0 2

 

Sample Output

2

 

Source

2013 ACM-ICPC南京赛区全国邀请赛——题目重现

 

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define maxn 5005
int pre[maxn];
int low[maxn];
vector<int>g[maxn];
int cut[maxn];
int dfsclock;
int tarjan(int u,int fa,int forbid)
{
    int lowu=pre[u]=++dfsclock;
    int child=0;
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        if(v==fa||v==forbid)
            continue;
        if(!pre[v])
        {
            child++;
            int lowv=tarjan(v,u,forbid);
            lowu=min(lowu,lowv);
            if(lowv>=pre[u])
            {
                cut[u]++;
            }
        }
        else
            lowu=min(lowu,pre[v]);
    }
    if(fa<0)
        cut[u]--;
    return low[u]=lowu;
}
int n,m;
int main()
{while(~scanf("%d%d",&n,&m))
{
    for(int i=0;i<n;i++)
        g[i].clear();
        for(int i=0;i<m;i++)


        {int u,v;
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        g[v].push_back(u);

        }
        int ans=-100;
        for(int u=0;u<n;u++)
        {
            int sum=0;
            dfsclock=0;
            memset(pre,0,sizeof(pre));
            memset(cut,0,sizeof(cut));
            for(int v=0;v<n;v++)
                if(v!=u&&!pre[v])
            {
                sum++;
            tarjan(v,-1,u);
            }
            for(int v=0;v<n;v++)
                if(v!=u)
                ans=max(ans,sum+cut[v]);
        }
        printf("%d\n",ans);

}
return 0;

}