HDU 4587 TWO NODES 删除两个点求连通分量

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最新推荐文章于 2020-08-27 00:36:55 发布

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本文介绍了一道关于无向图的算法题TWONODES，任务是计算删除两个节点后图中最大可能的连通分量数目。通过枚举第一个删除的节点，并运用Tarjan算法计算剩余节点删除后的连通分量个数，最终得出最优解。

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TWO NODES

Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 649 Accepted Submission(s): 203

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

Sample Output

Source

2013 ACM-ICPC南京赛区全国邀请赛——题目重现

让你删除两个点，求最多有多少个连通分量。
枚举要删除的第一个点，然后根据tarjan算法求剩下的每一个点被删除后的连通分量个数。如果一个连通分量只有一个点的话，删除此点此连通分量也就没有了。

//4843MS	420K
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define M 5007
using namespace std;
int low[M],dfn[M],head[M],cut[M];
bool vis[M];
int cnt,num,n,m;
struct E
{
    int v,next;
}edg[M*20];

void addedge(int u,int v)
{
    edg[cnt].v=v;
    edg[cnt].next=head[u];
    head[u]=cnt++;
}
void tarjan(int u,int fa)
{
    low[u]=dfn[u]=++num;
    vis[u]=true;
    for(int i=head[u];i!=-1;i=edg[i].next)
    {
        int v=edg[i].v;
        if(v==fa)continue;
        if(!dfn[v])
        {
            tarjan(v,fa);
            if(low[u]>low[v])
                low[u]=low[v];
            if(low[v]>=dfn[u])
                cut[u]++;//删除u点可以得到cut[u]个连通分量
        }
        else if(low[u]>dfn[v])
            low[u]=dfn[v];
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int u,v;
        memset(head,-1,sizeof(head));
        cnt=0;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        int ans=-1,sum;
        for(int i=0;i<n;i++)
        {
            memset(dfn,0,sizeof(dfn));
            memset(low,0,sizeof(low));
            memset(vis,false,sizeof(vis));
            for(int j=0;j<n;j++)cut[j]=1;
            num=sum=0;//sum代表连通分量的个数
            for(int j=0;j<n;j++)
            {
                if(i!=j&&!vis[j])
                {
                    sum++;cut[j]=0;
                    tarjan(j,i);
                }
            }
            for(int j=0;j<n;j++)
                if(j!=i)ans=max(ans,sum+cut[j]-1);
        }
        printf("%d\n",ans);
    }
    return 0;
}