之前只是听说过双连通分量,没有自己去写过,这次遇到了果断没有做出来,看题的时候算法是想到了的,奈何实在太挫了,写不出来。之前想用dfs找环再标记缩点,结果wa了,不知道怎么调就放弃了。今天看了一下Tarjan找桥和割点的方法还是比较简单的,当dfn[u]<low[v]时,就表示v顶点不经过父子边能到达的最远的点dfs编号比u小,就说明没有其他路可以到达u,故u->v是桥,然后标记一下,重新建图,求一个直径就好了。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cstdio>
#include <stack>
#include <queue>
#include <vector>
using namespace std;
#define MAXM 2000005
#define MAXN 200005
struct Edge
{
int v,next;
bool inbridge;
}edge[MAXM];
int head[MAXN],e;
void addedge(int u,int v)
{
edge[e].v=v;
edge[e].next=head[u];
edge[e].inbridge=false;
head[u]=e++;
edge[e].v=u;
edge[e].next=head[v];
edge[e].inbridge=false;
head[v]=e++;
}
int dfn[MAXN],low[MAXN],sign,bridge;
int instack[MAXN],scnt,belong[MAXN];
stack<int> s;
vector<int> g[MAXN];
void tarjan(int u,int p)
{
s.push(u);
dfn[u]=low[u]=++sign;
bool flag=true;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v==p&&flag)
{
flag=false;
continue;
}
if(!dfn[v])
{
tarjan(v,u);
if(low[u]>low[v])
low[u]=low[v];
if(dfn[u]<low[v])
{
bridge++;
edge[i].inbridge=true;
edge[i^1].inbridge=true;
}
}
else if(low[u]>dfn[v])
low[u]=dfn[v];
}
if(dfn[u]==low[u])
{
int i;
scnt++;
do
{
i=s.top();s.pop();
instack[i]=false;
belong[i]=scnt;
}
while(i!=u);
}
}
void work()
{
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(instack,0,sizeof(instack));
sign=scnt=bridge=0;
tarjan(1,0);
}
struct Node
{
int v,dis;
Node(int a,int b){v=a,dis=b;}
};
bool vis[MAXN];
int maxlen;
int bfs(int k)
{
memset(vis,0,sizeof(vis));
queue<Node> q;
q.push(Node(k,0));
int len=0,node=1;
vis[k]=true;
while(!q.empty())
{
Node u=q.front();
q.pop();
for(int i=0;i<g[u.v].size();i++)
{
int v=g[u.v][i];
if(!vis[v])
{
vis[v]=true;
if(len<u.dis+1)
{
len=u.dis+1;
node=v;
}
q.push(Node(v,u.dis+1));
}
}
}
maxlen=len;
return node;
}
int main()
{
//freopen("input.txt","r",stdin);
int n,m;
while(scanf("%d%d",&n,&m),n+m)
{
memset(head,-1,sizeof(head));
e=0;
for(int i=1;i<=n;i++)
g[i].clear();
for(int i=0;i<m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
addedge(u,v);
}
work();
for(int i=1;i<=n;i++)
{
for(int j=head[i];j!=-1;j=edge[j].next)
{
if(edge[j].inbridge)
{
g[belong[i]].push_back(belong[edge[j].v]);
}
}
}
int u=bfs(1);
int v=bfs(u);
printf("%d\n",bridge-maxlen);
}
}