# Network

 Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 10371 Accepted: 3853

### 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 in the network after each new link is added.

### 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

### Source

DFS搜索樹：用DFS對圖進行遍歷時，按照遍歷次序的不同，我們可以得到一棵DFS搜索樹。

id     123456
dfn   123456
low   111444

```  1 //2017-08-20
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6
7 using namespace std;
8
9 const int N = 100010;
10 const int M = 200010;
12 struct Edge{
13     int to, next;
14 }edge[M<<1];
15
16 void init(){
17     tot = 0;
19 }
20
21 void add_edge(int u, int v){
22     edge[tot].to = v;
25
26     edge[tot].to = u;
29 }
30
31 int n, m, deep, ans;
32 int dfn[N];//dfn[u]記錄節點u在DFS過程中被遍歷到的次序號
33 int low[N]; //low[u]記錄節點u或u的子樹通過非父子邊追溯到最早的祖先節點(即DFS次序號最小)
34 int fa[N];//保存dfs樹的信息
35 int level[N];//記錄節點在dfs樹中的深度
36 int bridge[N];//記錄割邊，若bridge[u] == 1, 則<u, fa[u]>為一條割邊
37
38 void tarjan(int u, int father){
39     fa[u] = father;
40     dfn[u] = low[u] = deep++;
41     level[u] = level[father]+1;
42     for(int i = head[u]; i != -1; i = edge[i].next){
43         int v = edge[i].to;
44         if(dfn[v] == -1){
45             tarjan(v, u);
46             low[u] = min(low[u], low[v]);
47             if(low[v] > dfn[u]){
48                 bridge[v] = 1;
49                 ans++;
50             }
51         }else if(v != father)
52               low[u] = min(low[u], dfn[v]);
53     }
54 }
55
56 void lca(int a, int b){
57     while(level[a] > level[b]){
58         if(bridge[a]){
59             ans--;
60             bridge[a] = 0;
61         }
62         a = fa[a];
63     }
64     while(level[b] > level[a]){
65         if(bridge[b]){
66             ans--;
67             bridge[b] = 0;
68         }
69         b = fa[b];
70     }
71     while(a != b){
72         if(bridge[a]){
73             ans--;
74             bridge[a] = 0;
75         }
76         if(bridge[b]){
77             ans--;
78             bridge[b] = 0;
79         }
80         a = fa[a];
81         b = fa[b];
82     }
83 }
84
85 int main()
86 {
87     //freopen("inputD.txt", "r", stdin);
88     int kase = 0;
89     while(scanf("%d%d", &n, &m)!=EOF && (n || m)){
90         printf("Case %d:\n", ++kase);
91         init();
92         int u, v;
93         for(int i = 0; i < m; i++){
94             scanf("%d%d", &u, &v);
96         }
97         memset(bridge, 0, sizeof(bridge));
98         memset(dfn, -1, sizeof(dfn));
99         memset(low, 0, sizeof(low));
100         level[0] = 0;
101         deep = 0;
102         tarjan(1, 0);
103         int q;
104         scanf("%d", &q);
105         while(q--){
106             scanf("%d%d", &u, &v);
107             if(ans)lca(u, v);
108             printf("%d\n", ans);
109         }
110         printf("\n");
111     }
112
113     return 0;
114 }```