I - 乓 (BFS+邻接表)

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).

InputThe input consists of multiple test cases. Each test case starts with a line containing two integers N and M (1 ≤ N ≤ 10,000, 0 ≤ M ≤ 1,000,000), which are the number of computers and the number of damaged links in USTC campus network, respectively. The computers are numbered from 1 to N and computer 1 is the BBS server. 
Each of the following M lines contains two integers A and B(1 ≤ A ≤ N, 1 ≤ B ≤ N, A ≠ B), which means the link between computer A and B is damaged. A link will appear at most once.

The last test case is followed by a line containing two zeros.OutputFor each test case, print a line containing the test case number( beginning with 1) followed by the number of computers still connecting with the BBS server.Sample Input

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

Sample Output

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 ;
}