LightOJ 1063 Ant Hills

Ant Hills

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on LightOJ. Original ID: 1063
64-bit integer IO format: %lld      Java class name: Main

After many years of peace, an ant-war has broken out.

In the days leading up to the outbreak of war, the ant government devoted a great deal of resources toward gathering intelligence on ant hills. It discovered the following:

1. The ant empire has a large network of ant-hills connected by bidirectional tracks.
2. It is possible to send a message from any ant hill to any other ant hill.

Now you want to stop the war. Since they sometimes attack your house and disturb you quite a lot. So, you have made a plan. You have a gun which can destroy exactly one ant-hill. So, you want to hit an ant hill if it can stop at least two other ant hills passing messages between them. Now you want the total number of ant hills you may choose to fire.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each test case contains a blank line and two integers n (1 ≤ n ≤ 10000), m (1 ≤ m ≤ 20000). n denotes the number of ant hills and m denotes the number of bi-directional tracks. Each of the next m lines will contain two different integers a b (1 ≤ a, b ≤ n) denoting that there is a track between a and b.

Output

For each case, print the case number and the total number of ant hills you may choose to fire.

Sample Input

2

5 4

2 1

1 3

5 4

4 1

3 3

1 2

2 3

1 3

Sample Output

Case 1: 2

Case 2: 0

Source

Problem Setter: Jane Alam Jan

解题：求割点数

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 10010;
 4 struct arc {
 5     int to,next;
 6     arc(int x = 0,int y = -1) {
 7         to = x;
 8         next = y;
 9     }
10 } e[100000];
11 int head[maxn],dfn[maxn],low[maxn];
12 int tot,idx,ans,n,m;
13 bool vis[maxn];
14 void add(int u,int v) {
15     e[tot] = arc(v,head[u]);
16     head[u] = tot++;
17     e[tot] = arc(u,head[v]);
18     head[v] = tot++;
19 }
20 void tarjan(int u,int fa) {
21     dfn[u] = low[u] = ++idx;
22     int son = 0;
23     for(int i = head[u]; ~i; i = e[i].next) {
24         if(!dfn[e[i].to]) {
25             tarjan(e[i].to,u);
26             low[u] = min(low[u],low[e[i].to]);
27             son++;
28             if(!vis[u]&&(fa == -1 && son > 1 || fa != -1 && low[e[i].to] >= dfn[u])) {
29                 vis[u] = true;
30                 ans++;
31             }
32         } else low[u] = min(low[u],dfn[e[i].to]);
33     }
34 }
35 int main() {
36     int T,u,v,cs = 1;
37     scanf("%d",&T);
38     while(T--) {
39         scanf("%d %d",&n,&m);
40         memset(head,-1,sizeof(head));
41         memset(dfn,0,sizeof(dfn));
42         memset(low,0,sizeof(low));
43         memset(vis,false,sizeof(vis));
44         ans = tot = 0;
45         for(int i = 0; i < m; ++i) {
46             scanf("%d %d",&u,&v);
47             add(u,v);
48         }
49         tarjan(1,-1);
50         printf("Case %d: %d\n",cs++,ans);
51     }
52     return 0;
53 }