在给出的两个点上加一条边,求剩下桥的数量,,不会LCA在线,就用了最普通的,先Tarjan双联通缩点,然后将缩完的图建成一棵树,树的所有边就是桥了,如果在任意两点间加一条边的话,那么从两点到最近公共祖先的所有边都不是桥了。。。。。。
#pragma comment(linker, "/STACK:10240000000000,10240000000000")
#include<stdio.h>
#include<stack>
#include<string.h>
#define N 100010
using namespace std;
int belong[N],low[N],dfs[N],head[N],num,n,idx,ans,nume,vis[N];
struct edge
{
int st,ed,next;
}E[N*4],e[N*4];
void addedge(int x,int y)
{
E[num].st=x;
E[num].ed=y;
E[num].next=head[x];
head[x]=num++;
}
void Addedge(int x,int y)
{
e[nume].st=x;
e[nume].ed=y;
e[nume].next=head[x];
head[x]=nume++;
}
stack<int>Q;
void Tarjan(int u,int father)
{
int i,v,flag=0;
low[u]=dfs[u]=idx++;
Q.push(u);
for(i=head[u];i!=-1;i=E[i].next)
{
v=E[i].ed;
if(dfs[v]==-1)
{
Tarjan(v,u);
low[u]=low[u]>low[v]?low[v]:low[u];
}
else if(v==father)
{
if(flag)
low[u]=low[u]>dfs[v]?dfs[v]:low[u];
flag++;
}
else low[u]=low[u]>dfs[v]?dfs[v]:low[u];
}
if(low[u]==dfs[u])
{
do
{
v=Q.top();
Q.pop();
belong[v]=ans;
}while(v!=u);
ans++;
}
}
void dfs1(int u)
{
vis[u]=1;
int i,v;
for(i=head[u];i!=-1;i=e[i].next)
{
v=e[i].ed;
if(vis[v]==1)continue;
dfs[v]=dfs[u]+1;
low[v]=u;
dfs1(v);
}
}
void Lca(int x,int y)
{
int i;
if(dfs[x]<dfs[y])
{i=x;x=y;y=i;}
while(dfs[x]>dfs[y])
{
if(vis[x]==0)
{
vis[x]=1;
ans--;
}
x=low[x];
}
while(x!=y)
{
if(vis[x]==0)
{
vis[x]=1;
ans--;
}
if(vis[y]==0)
{
vis[y]=1;
ans--;
}
x=low[x];y=low[y];
}
}
int main()
{
int i,k,x,y,m,op=1;
while(scanf("%d%d",&n,&m),n||m)
{
memset(head,-1,sizeof(head));
num=0;
for(i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
addedge(x,y);
addedge(y,x);
}
memset(dfs,-1,sizeof(dfs));
idx=ans=0;
Tarjan(1,-1);
memset(head,-1,sizeof(head));
nume=0;
for(i=0;i<num;i++)
{
x=belong[E[i].st];
y=belong[E[i].ed];
if(x==y)continue;
Addedge(x,y);
Addedge(y,x);
}
memset(vis,0,sizeof(vis));
dfs[0]=0;
low[0]=0;
dfs1(0);
memset(vis,0,sizeof(vis));
printf("Case %d:\n",op++);
scanf("%d",&k);
ans--;
while(k--)
{
scanf("%d%d",&x,&y);
Lca(belong[x],belong[y]);
printf("%d\n",ans); }
printf("\n");
}
return 0;
}