4612 warm up tarjan+bfs求树的直径（重边的强连通通分量）忘了写了，今天总结想起来了。

问加一条边，最少可以剩下几个桥。

先双连通分量缩点，形成一颗树，然后求树的直径，就是减少的桥。

本题要处理重边的情况。

如果本来就两条重边，不能算是桥。

还会爆栈，只能C++交，手动加栈了

别人都是用的双连通分量，我直接无向图改成有向图搞得强连通水过。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <iostream>
 #include <vector>
 #include <queue>
 #include <string.h>
 #include <stdlib.h>
 #include <stdio.h>
 #include <stack>

 using namespace std;
 const int maxn = ;
 bool vis[];
 struct edge
 {
     int v,next;
     edge()
     {
         next = -;
     }
 }edges[maxn*-];
 struct ed
 {
     int u,v;
 }e[maxn*-];
 int dfn[],low[],belong[];
 bool inst[];
 int g[maxn];
 vector<int>ng[maxn];

 stack<int>st;
 int bcnt,cnt,time;

 void init(int n)
 {
     int i;
     for(i =;i <= n;i++)
     g[i] = -;
     time = ;bcnt = cnt = ;
     return ;
 }
 void addedge(int u,int v,int val)
 {
     struct edge o;
     edges[cnt].v = v;
     edges[cnt].next = g[u];
     g[u] = cnt;
     cnt++;

     return ;
 }
 void tarjan(int i)
 {
     int j;
     dfn[i] = low[i] = ++time;
     inst[i] = ;
     st.push(i);

     for(j = g[i];j != -;j = edges[j].next)
     {
         if(vis[j]) continue;
         vis[j] = vis[j^] = ;
         int v;
         v = edges[j].v;
         if(!dfn[v])
         {
             tarjan(v);
             low[i] = min(low[i],low[v]);
         }
         else if(inst[v])
         low[i] = min(low[i],dfn[v]);
     }
     int k;
     if(dfn[i] == low[i])
     {

         bcnt++;
         do
         {
             k = st.top();
             st.pop();
             inst[k] = ;
             belong[k] = bcnt;

         }
         while(k != i);
     }

 }
 void tarjans(int n)
 {
     int i;
     bcnt = time = ;
     while(!st.empty())st.pop();
     memset(dfn,,sizeof(dfn));

     memset(inst,,sizeof(inst));
     memset(belong,,sizeof(belong));
     for(i = ;i <= n;i++)
     if(!dfn[i])tarjan(i);
 }
 struct node
 {
     int s,point;
 };

 struct node bfs(int s)
 {
     int i;
     for(i = ;i <= bcnt;i++)
     vis[i] = ;
     queue <struct node>q;
     struct node p;
     p.s = ;p.point = s;
     q.push(p);
     vis[s] = ;
     struct node max;
     max.s = ;

     while(!q.empty())
     {
         struct node now,temp;
         now = q.front();
         q.pop();

         for(i = ;i < ng[now.point].size();i++)
         {
             int v = ng[now.point][i];
             temp.s = now.s+;
             temp.point = v;
             if(!vis[v])
             {
                 vis[v] = ;
               //  cout<<v<<"***"<<endl;

                 if(max.s < temp.s)
                 max = temp;
                 q.push(temp);
             }
         }
     }
     return max;
 }
 int main()
 {
     int n,m;
     //freopen("in.txt","r",stdin);
     while(scanf("%d %d",&n,&m)&&(n||m))
     {
         int a,b,i;
         init(n);
         memset(vis,,sizeof(vis));
         for(i = ;i < m;i++)
         {
             scanf("%d %d",&e[i].u,&e[i].v);
             addedge(e[i].u,e[i].v,);
             addedge(e[i].v,e[i].u,);
         }
         tarjans(n);

         for(i = ;i <= n;i++)
         {
             ng[i].clear();
         }

         for(i = ;i < m;i++)
         {
             if(belong[e[i].u] == belong[e[i].v])
             continue;
             ng[belong[e[i].u]].push_back(belong[e[i].v]);
             ng[belong[e[i].v]].push_back(belong[e[i].u]);
         }
         struct node max;
         max = bfs();
         max = bfs(max.point);
         printf("%d\n",bcnt-max.s-);
     }
     return ;
 }