# Description

A network administrator manages a large network. The network consists
of N computers and M links between pairs of computers. Any pair of
computers are connected directly or indirectly by successive links, so
data can be transformed between any two computers. The administrator
finds that some links are vital to the network, because failure of any
one of them can cause that data can’t be transformed between some
computers. He call such a link a bridge. He is planning to add some
new links one by one to eliminate all bridges.

You are to help the administrator by reporting the number of bridges

# Input

The input consists of multiple test cases. Each test case starts with
a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤
200,000). Each of the following M lines contains two integers A and B
( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B.
Computers are numbered from 1 to N. It is guaranteed that any two
computers are connected in the initial network. The next line contains
a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links
the administrator plans to add to the network one by one. The i-th
line of the following Q lines contains two integer A and B (1 ≤ A ≠ B
≤ N), which is the i-th added new link connecting computer A and B.

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

# Output

For each test case, print a line containing the test case number(
beginning with 1) and Q lines, the i-th of which contains a integer
indicating the number of bridges in the network after the first i new
links are added. Print a blank line after the output for each test
case.

# Sample Input

``````3 2
1 2
2 3
2
1 2
1 3
4 4
1 2
2 1
2 3
1 4
2
1 2
3 4
0 0
``````

# Sample Output

``````Case 1:
1
0

Case 2:
2
0
``````

# 代码

``````#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
#define inf 1000000
#define mem(a,b) memset(a,b,sizeof(a))
const int N=100000+7;
const int M=400000+5;
int dfn[N],low[N],tot,times;
int first[N],pre[N];
int bridge[N],cnt;
int n,m,k,q=1;
struct edge
{
int v,next;
} e[M];
{
e[tot].v=v;
e[tot].next=first[u];
first[u]=tot++;
}
void init()
{
mem(first,-1);
mem(dfn,0);
mem(low,0);
mem(bridge,0);
mem(pre,0);
tot=0;
cnt=0;
times=0;
}
void tarjan(int u,int fa)
{
low[u]=dfn[u]=++times;
int flag=0;
for(int i=first[u]; ~i; i=e[i].next)
{
int v=e[i].v;
if(v==fa&&!flag)
{
flag=1;
continue;
}
if(!dfn[v])
{
pre[v]=u;
tarjan(v,u);
low[u]=min(low[u],low[v]);
if(low[v]>dfn[u])
{
cnt++;
bridge[v]=1;
}
}
else
low[u]=min(low[u],dfn[v]);
}
}

void lca(int u,int v)
{
while(dfn[u]>dfn[v])
{
if(bridge[u])
{
cnt--;
bridge[u]=0;
}
u=pre[u];
}
while(dfn[v]>dfn[u])
{
if(bridge[v])
{
cnt--;
bridge[v]=0;
}
v=pre[v];
}
while(u!=v)
{
if(bridge[u])
{
cnt--;
bridge[u]=0;
}
if(bridge[v])
{
cnt--;
bridge[v]=0;
}
u=pre[u];
v=pre[v];
}
}

int main()
{
int u,v;
while(scanf("%d%d",&n,&m)&&(n||m))
{
init();
for(int i=1; i<=m; i++)
{
scanf("%d%d",&u,&v);
}
for(int i=1; i<=n; i++)
if(!dfn[i])
tarjan(i,-1);
printf("Case %d:\n",q++);
scanf("%d",&k);
while(k--)
{
scanf("%d%d",&u,&v);
lca(u,v);
printf("%d\n",cnt);
}
puts("");
}
return 0;
}

``````