学步园

tarjan缩点+树的直径

https://www.byvoid.com/blog/scc-tarjan/    讲tarjian比较好

http://blog.csdn.net/u010638776/article/details/9472605   借鉴了这里

#include<iostream>
#include<cstdio>
#include<cstdio>
#include<queue>
#pragma comment(linker, "/STACK:1024000000,1024000000")   //扩栈 
using namespace std;
#define maxn 210000
#define maxm 2100000
int n,m;
int head[maxn],v[maxm],next[maxm],cnt;
int head1[maxn],v1[maxm],next1[maxm],cnt1;
int dfn[maxn],low[maxn],step;
int sta[maxn],top;
int ID[maxn],IdNum;
int vis[maxn];
int r;
void Init() 
{
	memset(vis,0,sizeof(vis));
	memset(head1,-1,sizeof(head1));
	memset(head,-1,sizeof(head));
	memset(dfn,0,sizeof(dfn));
	memset(low,0,sizeof(low));
	IdNum=top=step=cnt1=cnt=0;
}
void add1(int a,int b)
{
	v1[cnt1]=b;
	next1[cnt1]=head1[a];
	head1[a]=cnt1++;
}
void add(int a,int b)
{
	v[cnt]=b;
	next[cnt]=head[a];
	head[a]=cnt++;
}
void tarjan(int u,int pre) //pre为前一个点 
{
	dfn[u]=low[u]=++step;
	sta[++top]=u;
	vis[u]=1;
	int k=0;
	for(int i=head[u];~i;i=next[i])
	{
		int to=v[i];
		if(!dfn[to])//如果没有进过栈 
		{
			tarjan(to,u);//继续找 
			low[u]=min(low[u],low[to]);
		}
		else
		{
			if(to==pre)//如果是前一个点 
			{
				if(k)//考虑了重边的可能 
					low[u]=min(low[u],dfn[to]);
				k++;//无向边是建2条边 
			}
			else if(vis[to])
				low[u]=min(low[u],dfn[to]);
		}
	}
	if(dfn[u]==low[u])
	{
		IdNum++;
		int x;
		do
		{
			x=sta[top--];
			vis[x]=0;
			ID[x]=IdNum;//表示在同一个强连通分量 
		}while(u!=x);
	}
}
int bfs(int x)
{
	queue<int> q;
	int dis[maxn];
	int end;
	memset(dis,-1,sizeof(dis));
	q.push(x);
	dis[x]=0;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		r=dis[u];
		end=u;
		for(int i=head1[u];~i;i=next1[i])
		{
			int to=v1[i];
			if(dis[to]==-1)
			{
				dis[to]=dis[u]+1;
				q.push(to);
			}
		}
	}
	return end;
}
int main()
{
	int x,y;
	while(~scanf("%d%d",&n,&m)&&n+m)
	{
		Init();
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&x,&y);
			if(x==y)
				continue;
			add(x,y);
			add(y,x);
		}	
		tarjan(1,-1);
	//	cout<<IdNum<<endl;
		for(int i=1;i<=n;i++)// 缩点建边 
			for(int j=head[i];~j;j=next[j])
			{
				int to=v[j];
				if(ID[i]!=ID[to])
					add1(ID[i],ID[to]);
			}
		bfs(bfs(1));//2次bfs找树的直径 
		printf("%d\n",IdNum-1-r);
	}
	return 0;
}