在无向图求完tarjan 并且缩点后 剩下的是树 树上的边全是 割边 添加 x，y联通块 之间的树边全变为普通边 所以我们只需要 维护联通块 即可 可以使用并查集 复杂度约为 mq 此题我们可以直接考虑 用 on的方法来找lca 复杂度为 nm 考虑一下正确性
考虑一下此图 点2到点5与点4到点7是否会错过 因为 倘若先走2 - 5那么6 就会成为7的父节点 而不是 4 若先走2 - 3那么 也是正确 所以on找 父节点是可以的

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

const int N = 1e5 + 10,M = 2e5 + 10;
int n,m;
int head[N],to[M * 2],last[M * 2],cnt;
void add(int a,int b){
    
    
    to[++cnt] = b;
    last[cnt] = head[a];
    head[a] = cnt;
}

int dfn[N],low[N],times,top,pre[N],bridge[N],sum;
void tarjan(int x,int lastt){
    
    
    dfn[x] = low[x] = ++times;
    for(int i = head[x]; i != -1; i = last[i]){
    
    
        int j = to[i];
        if(!dfn[j]){
    
    
            tarjan(j,i);
            low[x] = min(low[x],low[j]);
            pre[j] = x;
            if(dfn[x] < low[j]){
    
    
                bridge[j] = 1;
                sum++;
            }
        }else if(i != (lastt ^ 1)){
    
    
            low[x] = min(low[x],dfn[j]);
        }
    }
}

int judge(int x,int y){
    
    
    int ans = 0;
    while(dfn[x] > dfn[y]){
    
    
        if(bridge[x]){
    
    
            ans++;
            bridge[x] = 0;
        }
        x = pre[x];
    }

    while(dfn[x] < dfn[y]){
    
    
        if(bridge[y]){
    
    
            ans++;
            bridge[y] = 0;
        }
        y = pre[y];
    }
    while(x != y){
    
    
        if(bridge[x]){
    
    
            bridge[x] = 0;
            ans++;
        }
        if(bridge[y]){
    
    
            bridge[y] = 0;
            ans++;
        }
        x = pre[x];
        y = pre[y];
    }
    return ans;
}

int main(){
    
    
    int CASE = 0;
    while(~scanf("%d%d",&n,&m)){
    
    
        if(n == 0 && m == 0) break;
        cnt = 1;
        sum = 0;
        memset(head,-1,sizeof head);
        memset(dfn,0,sizeof dfn);
        memset(low,0,sizeof low);
        memset(bridge,0,sizeof bridge);
        for(int i = 1; i <= m; i++){
    
    
            int x,y;
            scanf("%d%d",&x,&y);
            add(x,y);
            add(y,x);
        }
        for(int i = 1; i <= n; i++){
    
    
            if(!dfn[i]){
    
    
                tarjan(i,0);
            }
        }
        printf("Case %d:\n",++CASE);
        int q;
        scanf("%d",&q);
        while(q--){
    
    
            int x,y;
            scanf("%d%d",&x,&y);
            cout << (sum -= judge(x,y)) << endl;
        }

    }





    return 0;
}