hdu4587 求割点变形

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http://acm.hdu.edu.cn/showproblem.php?pid=4587

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.

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

2

/**
hdu4587 求割点变形
题目大意：给定一个图，去掉其中哪两点后能得到最多的连通块，是多少
解题思路：枚举去掉第一个点，然后利用求割点的模板就可以求出去掉另一点后的答案，取最多即可
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=5010;

struct note
{
    int v,next;
    bool cut;
} edge[maxn*2];

int head[maxn],ip;

void init()
{
    memset(head,-1,sizeof(head));
    ip=0;
}

int low[maxn],dfn[maxn],st[maxn],dex,top;
bool inst[maxn],cut[maxn];
int add_block[maxn];
int bridge;
int n,m,now;

void addedge(int u,int v)
{
    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;
}

void tarjan(int u,int pre)
{
    low[u]=dfn[u]=++dex;
    st[top++]=u;
    inst[u]=true;
    int son=0;
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==pre)continue;
        if(v==now)continue;
        if(!dfn[v])
        {
            son++;
            tarjan(v,u);
            if(low[u]>low[v])low[u]=low[v];
            if(low[v]>dfn[u])
            {
                bridge++;
                edge[i].cut=true;
                edge[i^1].cut=true;
            }
            if(u!=pre&&low[v]>=dfn[u])
            {
                cut[u]=true;
                add_block[u]++;
            }
        }
        else if(low[u]>dfn[v])
        {
            low[u]=dfn[v];
        }
    }
    if(u==pre&&son>1)cut[u]=true;
    if(u==pre)add_block[u]=son-1;
    inst[u]=false;
    top--;
}

int solve()
{
    memset(dfn,0,sizeof(dfn));
    memset(inst,false,sizeof(inst));
    memset(add_block,0,sizeof(add_block));
    memset(cut,false,sizeof(cut));
    dex=top=0;
    bridge=0;
    int cnt=0;
    for(int i=0;i<n;i++)
    {
        if(i!=now&&!dfn[i])
        {
            cnt++;
            tarjan(i,i);
        }
    }
    int ans=-1;
    for(int i=0;i<n;i++)
    {
        if(i!=now)
        {
           ans=max(ans,cnt+add_block[i]);
        }
    }
    return ans;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        int ans=-1;
        for(int i=0;i<n;i++)
        {
            now=i;
            ans=max(ans,solve());
        }
        printf("%d\n",ans);
    }
    return 0;
}

hdu4587 求割点变形

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原文地址：http://blog.csdn.net/lvshubao1314/article/details/44025035