HDU 4587 TWO NODES 枚举+割点

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

TWO NODES

Time Limit: 24000/12000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1448 Accepted Submission(s): 441


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each graph in the input, you should output the value of stab.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

2

[align=left]Source[/align]
2013 ACM-ICPC南京赛区全国邀请赛——题目重现

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题意

给你个图，问你去掉两个点之后能有最多多少连通块。

题解

先枚举其中一个点，然后在剩下的点中求割点，Tarjan的时候统计一下每个割点分割几个连通块，取个最大的割点，然后再dfs一次求连通块个数。

代码

#include<cstdio>
#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#define MAX_N 5555
using namespace std;

vector<int> G[MAX_N];
bool vis[MAX_N];
int dfn[MAX_N],low[MAX_N],ind=0;

int cut[MAX_N];

int node;

void Tarjan(int u,int p){
int child=0;
dfn[u]=low[u]=++ind;
vis[u]=1;
for(int i=0;i<G[u].size();i++){
int v=G[u][i];
if(v==p||v==node)continue;
if(!vis[v]){
Tarjan(v,u);
low[u]=min(low[v],low[u]);
child++;
if((p==-1&&child>1)||(p!=-1&&low[v]>=dfn[u]))
cut[u]++;
}
else
low[u]=min(dfn[v],low[u]);
}
}

int n,m;

void init(){
for(int i=0;i<=n;i++)G[i].clear();
ind=0;
memset(vis,0,sizeof(vis));
memset(cut,0,sizeof(cut));
}

bool used[MAX_N];
int cu;
void dfs(int u,int p){
if(u==p||used[u]||u==node||u==cu)return;
used[u]=1;
for(int i=0;i<G[u].size();i++)dfs(G[u][i],u);
}

int main(){
while(scanf("%d%d",&n,&m)==2){
int stab=1;
init();
int u,v;
for(int i=0;i<m;i++) {
scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
for(int i=0;i<n;i++){
node=i;
memset(vis,0,sizeof(vis));
ind=0;
memset(cut,0,sizeof(cut));
for(int j=0;j<n;j++)
if((!vis[j])&&j!=node)
Tarjan(j,-1);
int maxC=0;
for(int j=0;j<n;j++)
if(j!=node&&cut[j]>=maxC){
cu=j;
maxC=cut[j];
}
int ans=0;
memset(used,0,sizeof(used));
for(int j=0;j<n;j++)
if((!used[j])&&j!=node&&j!=cu){
dfs(j,-1);
ans++;
}
stab=max(stab,ans);
}
printf("%d\n",stab);
}

return 0;
}