http://acm.nbut.cn/Contest/view/id/43/problem/B.xhtml

• 问题描述
• Death-Moon loves telling stories.
  Some days ago, he told us a funny story.

   

  Long long ago, there is a hamster who is so naughty. Now, he comes to a place likes a N * N square. so, he is so excited to drill holes underground.

  • 输入
  • Input until EOF.
    Fisrt input two integers N (3 <= N < 100) and M (0 < M <2000). N is the side of N * N ground, M is the number of operations.
    
    Then follow M lines.
    Each line is a command:
    
    Out x y : Hamster will get out at (x, y) from underground. So, place (x, y) will be a hole.
    P : Calculate out the number of the holes and output (Two holes are connected when they share an edge).
    If two holes are connected, it means it is one hole.
  • 输出
  • For each 'P' command, output the number of the holes. Maybe hamster will get out at the same place more than once

  思路，用hash和并查积

  •  

  代码：

•  

   

  #include <cstdio>
  #include <set>
  #include <cstring>
  using namespace std;
  #define N 101
  int f[N * N * N];
  int a[N][N];
  set<int> v;
  int ff(int x)
  {
      if(f[x] == x)
      return x;
      return ff(f[x]);
  }
  int main()
  {
      int n,m;
      while(~scanf("%d %d",&n,&m))
      {
          memset(a,0,sizeof(a));
          char s[10];
          v.clear();
          for(int i = 1;i <= n;i ++)
          for(int j = 1;j <= n;j ++)
          f[i * 10000 + j] = i * 10000 + j;
          while(m --)
          {
             scanf("%s",s);
             if(s[0] == 'P')
             printf("%d\n",v.size());
             else
             {
                 int x,y,p,q;
                 scanf("%d %d",&x,&y);
                 x ++;
                 y ++;
                 q = ff(x * 10000 + y);
                 a[x][y] = 1;
                 set<int>:: iterator cp;
                 if(!a[x - 1][y] && !a[x + 1][y] && !a[x][y - 1] && !a[x][y + 1])
                 {
                     v.insert(x * 10000 + y);
                 }
                if(a[x - 1][y])
                 {
                     int u = (x - 1) * 10000 + y;
                     p = ff(u);
                     f[p] = q;
                     cp = v.find(p);
                     if(cp != v.end())
                     v.erase(cp);
                     v.insert(q);
                  }
                if(a[x + 1][y])
                  {
                      int u = (x + 1) * 10000 + y;
                      p = ff(u);
                     f[p] = q;
                     cp = v.find(p);
                     if(cp != v.end())
                     v.erase(cp);
                     v.insert(q);
                  }
                  if(a[x][y - 1])
                  {
                      int u = x * 10000 + y - 1;
                      p = ff(u);
                      f[p] = q;
                     cp = v.find(p);
                     if(cp != v.end())
                     v.erase(cp);
                     v.insert(q);
                  }
                  if(a[x][y + 1])
                  {
                      int u = x * 10000 + y + 1;
                      p = ff(u);
                     f[p] = q;
                     cp = v.find(p);
                     if(cp != v.end())
                     v.erase(cp);
                     v.insert(q);
                  }
             }
          }
      }
      return 0;
  }