TWO NODES

Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

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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题目大意：一个无向图去掉两个点后，最多产生几个联通分量

思路：先去掉一个点，再对去掉点的图进行tarjan求割点

ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
struct s
{
	int u,v,next;
}edge[10010];
int head[5050],dfn[5050],low[5050],n,m,time,sum,ans,cnt,del,iscnt[5050];
void init()
{
	time=cnt=sum=ans=0;
	memset(head,-1,sizeof(head));
	//	memset(vis,0,sizeof(vis));
	memset(iscnt,0,sizeof(iscnt));
}
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void tarjan(int u,int pre)
{
	low[u]=dfn[u]=time++;
	int i,j;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if(v==del)
			continue;
		if(dfn[v]==-1)
		{
			tarjan(v,u);
			low[u]=min(low[v],low[u]);
			if(low[v]>=dfn[u])
				iscnt[u]++;
		}
		else
		{
			if(dfn[v]<dfn[u]&&v!=u)
			{
				low[u]=min(low[u],dfn[v]);
			}
		}
	}
	if(pre<0)
		iscnt[u]--;
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i,j;
		init();
		for(i=0;i<m;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			add(u,v);
			add(v,u);
		}
		for(i=0;i<n;i++)
		{
			del=i;
			time=0;
			memset(low,-1,sizeof(low));
			memset(dfn,-1,sizeof(dfn));
			memset(iscnt,0,sizeof(iscnt));
			sum=0;
			for(j=0;j<n;j++)
			{
				if(j==del)
					continue;
				if(dfn[j]==-1)
				{
					tarjan(j,-1);
					sum++;
				}
			}
			for(j=0;j<n;j++)
			{
				if(j==del)
					continue;
			//	printf("%d %d %d \n",sum,j,iscnt[j]);
				ans=max(ans,sum+iscnt[j]);
			}
		}
		printf("%d\n",ans);
	}
}