You are to help the administrator by reporting the number of bridges in the network after each new link is added.

The last test case is followed by a line containing two zeros.

传送门：【HDU】2460 Network

题目分析：首先tarjan求出双连通分量，然后重新建边，dfs将无根树转化为有根树，记录父亲、节点深度，并且记录桥的数量。然后每次询问非递归的查找询问两点的最近公共祖先，根据深度平衡两端的抬升，直到遇到最近公共祖先（代码中则是 u == v ）抬升节点的同时，将路过的节点压栈，同时减去路过的边的数量（一定是桥），剩下桥的数量就是减去后的个数，然后对所有在栈中的节点全部指向最近公共祖先（等效于路径压缩）。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#pragma comment(linker,"/STACk:10240000,10240000")
using namespace std ;

#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clear( a , x ) memset ( a , x , sizeof a )

typedef long long LL ;

const int MAXN = 100005 ;
const int MAXE = 200005 ;

struct Edge {
	int v , n ;
	Edge ( int var = 0 , int next = 0 ) :
		v ( var ) , n ( next ) {}
} ;

struct BCC {
	Edge edge[MAXE << 1] ;
	int adj[MAXN] , cntE ;
	int bcc[MAXN] , bcc_cnt ;
	int ins[MAXN] , S[MAXN] , dfs_clock , top ;
	int low[MAXN] , dfn[MAXN] ;
	
	void init () {
		top = cntE = dfs_clock = bcc_cnt = 0 ;
		clear ( ins , 0 ) ;
		clear ( dfn , 0 ) ;
		clear ( adj , -1 ) ;
	}
	
	void addedge ( int u , int v ) {
		edge[cntE] = Edge ( v , adj[u] ) ;
		adj[u] = cntE ++ ;
		edge[cntE] = Edge ( u , adj[v] ) ;
		adj[v] = cntE ++ ;
	}
	
	void tarjan ( int u , int fa ) {
		low[u] = dfn[u] = ++ dfs_clock ;
		ins[u] = 1 ;
		S[top ++] = u ;
		int flag = 1 ;
		for ( int i = adj[u] ; ~i ; i = edge[i].n ) {
			int v = edge[i].v ;
			if ( v == fa && flag ) {
				flag = 0 ;
				continue ;
			}
			if ( !dfn[v] ) {
				tarjan ( v , u ) ;
				low[u] = min ( low[u] , low[v] ) ;
			}
			else if ( ins[v] )
				low[u] = min ( low[u] , dfn[v] ) ;
		}
		if ( low[u] == dfn[u] ) {
			++ bcc_cnt ;
			while ( 1 ) {
				int v = S[-- top] ;
				ins[v] = 0 ;
				bcc[v] = bcc_cnt ;
				if ( v == u )
					break ;
			}
		}
	}
	
	void find_bcc ( int n ) {
		REPF ( i , 1 , n )
			if ( !dfn[i] )
				tarjan ( i , -1 ) ;
	}
} ;

struct LCA {
	Edge edge[MAXE << 1] ;
	int adj[MAXN] , cntE ;
	int deep[MAXN] ;
	int p[MAXN] ;
	int pre[MAXN] ;
	int tot ;
	int S[MAXN] , top ;
	
	void init () {
		cntE = 0 ;
		clear ( adj , -1 ) ;
	}
	
	void addedge ( int u , int v ) {
		edge[cntE] = Edge ( v , adj[u] ) ;
		adj[u] = cntE ++ ;
		edge[cntE] = Edge ( u , adj[v] ) ;
		adj[v] = cntE ++ ;
	}
	
	int find ( int x ) {
		return p[x] == x ? x : ( p[x] = find ( p[x] ) ) ;
	}
	
	void dfs ( int u , int fa ) {
		pre[u] = u ;
		for ( int i = adj[u] ; ~i ; i = edge[i].n ) {
			int v = edge[i].v ;
			if ( v == fa )
				continue ;
			++ tot ;
			deep[v] = deep[u] + 1 ;
			dfs ( v , u ) ;
			pre[v] = u ;
		}
	}
	
	void preProcess ( int n ) {
		REPF ( i , 1 , n )
			p[i] = i ;
		tot = 0 ;
		deep[1] = 0 ;
		dfs ( 1 , 0 ) ;
	}
	
	void solve ( int u , int v ) {
		u = find ( u ) ;
		v = find ( v ) ;
		top = 0 ;
		while ( u != v ) {
			if ( deep[u] > deep[v] )
				S[top ++] = u , u = find ( pre[u] ) ;
			else
				S[top ++] = v , v = find ( pre[v] ) ;
			-- tot ;
		}
		while ( top )
			p[S[-- top]] = u ;
		printf ( "%d\n" , tot ) ;
	}
} ;	

BCC c ;
LCA a ;
int uu[MAXE] , vv[MAXE] ;

void work () {
	int cas = 0 ;
	int n , m ;
	int u , v ;
	while ( ~scanf ( "%d%d" , &n , &m ) && ( n || m ) ) {
		printf ( "Case %d:\n" , ++ cas ) ;
		c.init () ;
		a.init () ;
		REP ( i , m ) {
			scanf ( "%d%d" , &uu[i] , &vv[i] ) ;
			c.addedge ( uu[i] , vv[i] ) ;
		}
		c.find_bcc ( n ) ;
		REP ( i , m )
			if ( c.bcc[uu[i]] != c.bcc[vv[i]] )
				a.addedge ( c.bcc[uu[i]] , c.bcc[vv[i]] ) ;
		a.preProcess ( c.bcc_cnt ) ;
		scanf ( "%d" , &m ) ;
		while ( m -- ) {
			scanf ( "%d%d" , &u , &v ) ;
			a.solve ( c.bcc[u] , c.bcc[v] ) ;
		}
		printf ( "\n" ) ;
	}
}

int main () {
	work () ;
	return 0 ;
}