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 10010bool 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;}