HDU4612+Tarjan缩点+BFS求树的直径

tarjan+缩点+树的直径
题意：给出n个点和m条边的图，存在重边，问加一条边以后，剩下的桥的数量最少为多少。
先tarjan缩点，再在这棵树上求直径。加的边即是连接这条直径的两端。

 /*
 tarjan+缩点+树的直径
 题意：给出n个点和m条边的图，存在重边，问加一条边以后，剩下的桥的数量最少为多少。
 先tarjan缩点，再在这棵树上求直径。加的边即是连接这条直径的两端。
 */
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<stack>
 #include<algorithm>
 #include<iostream>
 #include<queue>
 #include<map>
 #include<math.h>
 using namespace std;
 typedef long long ll;
 //typedef __int64 int64;
 const int maxn = ;
 const int maxm = ;
 const int inf = 0x7fffffff;
 const double pi=acos(-1.0);
 const double eps = 1e-;
 struct Edge{
     int v,next;
 }edge[ maxm<< ],edgeTree[ maxm<< ];
 int cnt,cnt2,head[ maxn ],head2[ maxn ];
 int vis[ maxn ];
 int dfn[ maxn ];
 int low[ maxn ];
 int be[ maxn ];//缩点
 int sum;//the num of "缩点"
 int id;
 stack<int>sta;
 queue<int>q;
 int maxNode,maxs;//这棵树的直径maxs
 int dis[ maxn ];
 map<int,int>mp;
 void init(){
     cnt = cnt2 = ;
     id = ;
     sum = ;
     //mp.clear();
     memset( head,-,sizeof( head ) );
     memset( head2,-,sizeof( head2 ) );
     memset( dfn,-,sizeof( dfn ) );
     memset( vis,-,sizeof( vis ) );
     memset( low,-,sizeof( low ) );
     while( !sta.empty() )
         sta.pop();
 }
 void addedge( int a,int b ){
     edge[ cnt ].v = b;
     edge[ cnt ].next = head[ a ];
     head[ a ] = cnt++;
 }
 void addedge2( int a,int b ){
     edgeTree[ cnt2 ].v = b;
     edgeTree[ cnt2 ].next = head2[ a ];
     head2[ a ] = cnt2++;
 }

 void tarjan( int u,int Link ){
     dfn[ u ] = low[ u ] = id++;
     vis[ u ] = ;
     sta.push( u );
     int cc = ;
     for( int i=head[u];i!=-;i=edge[i].next ){
         int v = edge[ i ].v;
         if( dfn[ v ]==- ){
             tarjan( v,u );
             low[ u ] = min( low[ u ],low[ v ] );
         }
         else if( Link==v ){
             if( cc ) low[ u ] = min( low[ u ],dfn[ v ] );
             cc++;
         }
         else if( vis[ v ]== ){
             low[ u ] = min( low[ u ],dfn[ v ] );
         }
     }
     if( dfn[u]==low[u] ){
         sum++;
         int tt;
         while(  ){
             tt = sta.top();
             sta.pop();
             vis[ tt ] = ;
             be[ tt ] = sum;
             if( tt==u ) break;
         }
     }
 }
 void buildTree( int n ){
     for( int i=;i<=n;i++ ){
         for( int j=head[ i ];j!=-;j=edge[ j ].next ){
             if( be[i]!=be[edge[j].v] ){
                 addedge2( be[i],be[edge[j].v] );
             }
         }
     }
 }
 void bfs( int s,int n ){
     memset( vis,,sizeof( vis ) );
     vis[ s ] = ;
     while( !q.empty() )
         q.pop();
     q.push( s );
     dis[ s ] = ;
     maxs = ;
     while( !q.empty() ){
         int cur = q.front();
         q.pop();
         if( dis[ cur ]>=maxs ){
             maxs = dis[ cur ];
             maxNode = cur;
         }
         for( int i=head2[ cur ];i!=-;i=edgeTree[ i ].next ){
             int v = edgeTree[ i ].v;
             if( vis[ v ]== ) continue;
             vis[ v ] = ;
             dis[ v ] = dis[ cur ]+;
             q.push( v );
         }
     }
     return ;
 }
 int dfs(int u,int p)
 {
     int max1=,max2=;
     for (int i=head2[u];i!=-;i=edgeTree[i].next)
     {
         int v=edgeTree[i].v;
         if (v==p) continue;
         int tmp=dfs(v,u)+;
         if (max1<tmp) max2=max1,max1=tmp;
         else if (max2<tmp) max2=tmp;
     }
     maxs=max(maxs,max1+max2);
     return max1;
 } //dfs求树的直径 ok
 int main(){
     int n,m;
     while( scanf("%d%d",&n,&m)&&(n+m) ){
         int a,b;
         init();
         while( m-- ){
             scanf("%d%d",&a,&b);
             if( a==b ) continue;
             //if( mp[min(a,b)] == max(a,b) ) continue;
             //mp[min(a,b)] = max(a,b);
             //不能这样去重边？？？
             addedge( a,b );
             addedge( b,a );
         }
         for( int i=;i<=n;i++ ){
             if( dfn[i]==- ){
                 tarjan( i,- );
             }
         }
         buildTree( n );
         bfs( ,sum );
         bfs( maxNode,sum );
         //printf("sum=%d, maxs=%d\n",sum,maxs);
         //maxs = 0;
         //dfs( 1,-1 );
         printf("%d\n",sum--maxs);
     }
     return ;
 }