NOJ [1301] Gopher Hole

一开始我没想到，如果原来有两个洞无法合并，加入第三个洞如果可以合并，那么三个洞就会变1个洞，所以原来写的WA了，后来参照了下题解的方法，才AC了，果然太弱了啊

• 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. 
  
  

  #include<stdio.h>
  #include<string.h>
  #define maxn 10010
  bool is_break[maxn];
  int father[maxn];
  int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
  int is_not=0;
  int find(int x)
  {
    if(x==father[x])
      return father[x];
    if(x!=father[x])
      father[x]=find(father[x]);
    return father[x];
  }
  
  void Union(int x,int y)
  {
    int xx=find(x);
    int yy=find(y);
    if(xx!=yy)
    {
      father[yy]=xx;
      is_not--;
    }
  }
  
  int main()
  {
    int n,m,i,j;
    while(~scanf("%d%d",&n,&m))
    {
      for(i=0;i<=n*n+10;i++)
        father[i]=i;
      memset(is_break,false,sizeof(is_break));
      char com[4];
      int x,y,xx,yy,ans=0,dit;
      while(m--)
      {
        scanf("%s",com);
        if(com[0]=='O')
        {
          scanf("%d%d",&x,&y);
          dit=x*n+y;
          if(is_break[dit])
            continue;
          is_break[dit]=1;
          for(j=0;j<4;j++)
          {
            xx=x+dir[j][0];
            yy=y+dir[j][1];
            if(xx>=0 && yy>=0 && xx<n && yy<n)
            {
              if(is_break[xx*n+yy]==0)
               continue;
            else
               break;
            }
          }
          if(j==4)
          {
            ans++;
            continue;
          }
          is_not=1;
          for(j=0;j<4;j++)
          {
            xx=x+dir[j][0];
            yy=y+dir[j][1];
            if(xx<0 || yy<0 || xx>=n || yy>=n || is_break[xx*n+yy]==0)
              continue;
            else
              Union(dit,xx*n+yy);
          }
          ans+=is_not;
        }
        else if(com[0]=='P')
          printf("%d\n",ans );
      }
    }
    return 0;
  }