USTC campus network is a huge network. There is a bi-directional link between every pair of computers in the network. One of the computers is the BBS server, which is so popular that thousands of people log on it every day. Recently some links of the network are damaged by the rainstorm. The network administrator is going to check which computers are still connected with the BBS server directly or indirectly.

You are to help the administrator to report the number of computers still connecting with the BBS server (not including itself).

3 2

1 2

1 3

4 3

1 2

3 2

4 2

0 0

Case 1: 0

Case 2: 2

要用邻接表存一下不能走的地方，然后预处理一下，直接用二维的数组存会MLE，还有尽量用C++交，G++容易T
代码：

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

#include<map>

#include<vector>

#include<cmath>

#define Inf 0x3f3f3f3f

const int maxn=1e4+;

typedef long long ll;

using namespace std;

bool book[maxn];

bool vis[maxn];

vector<int>vec[maxn];

int n,m;

int bfs()

{

    queue<int>q;

    book[]=true;

    q.push();

    int ans=;

    while(!q.empty())

    {

    int tmp=q.front();

    q.pop();

    memset(vis,false,sizeof(vis));

    for(int t=;t<vec[tmp].size();t++)

    {

        vis[vec[tmp][t]]=true;

    }

    for(int t=;t<=n;t++)

    {

        if(vis[t]==false&&book[t]==false)

        {

            ans++;

            q.push(t);

            book[t]=true;

        }

    }

    }

    return ans;

}

int main()

{

    int cnt=;

       while(scanf("%d%d",&n,&m)!=EOF)

       {

           if(n==&&m==)

           {

               break;

           }

           for(int t=;t<=n;t++)

           {

               vec[t].clear();

           }

           memset(book,false,sizeof(book));

           int a,b;

           for(int t=;t<m;t++)

           {

               scanf("%d%d",&a,&b);

               vec[a].push_back(b);

               vec[b].push_back(a);

           }

           int ans=bfs();

           printf("Case %d: %d\n",cnt++,ans);

       }

    return ;

}