hdu 1198 Farm Irrigation(并查集)

http://acm.hdu.edu.cn/showproblem.php?pid=1198

原来看解题报告做过的一道题，现在拾起来再做感觉还是有点难度，慢慢想想也就思路清晰了。首先发现通道只要相碰就能连接。结构体记录是否能够连接，把能连接的用并查集归为同一集合，不能连接的则属于不同的集合。最后统计几何数就欧了。。、

#include <cstdio>
#include  <cstring>
#define MAXN 59
struct node
{
  int up;
  int down;
  int left;
  int right;
};
node  tagp[11]=
{
  {1,0,1,0},{1,0,0,1},
  {0,1,1,0},{0,1,0,1},
  {1,1,0,0},{0,0,1,1},
  {1,0,1,1},{1,1,1,0},
  {0,1,1,1},{1,1,0,1},
  {1,1,1,1}
};//记录是否连接情况1代表连接，0代表不行
int f[MAXN*MAXN];
void init()
{
    for (int i = 0; i < MAXN*MAXN;i++)
    {
        f[i] = i;
    }
}
int find(int x)
{
    if(x != f[x])
    f[x] = find(f[x]);
    return f[x];
}
void Union(int x,int y)
{
   x = find(x);
   y = find(y);
   if(x != y)
   f[y]=x;
}
int main()
{
   char str[MAXN][MAXN];
   int n,m,i,j;
   int x,y;
   while (scanf("%d%d",&n,&m))
   {
       init();
       if(n == -1&&m == -1)
       break;
       for (i = 0; i < n; ++i)
       scanf("%s",str[i]);
       int a,b,c;
//从头一直检查右边以及下边即可
       for (i = 0; i < n; ++i)
       {
           for (j = 0; j < m; ++j)
           {
               a = str[i][j] - 'A';
               b = str[i][j+1] - 'A';
               c = str[i+1][j] - 'A';
               if(j+1<m)//边界考虑
               {
                   if(tagp[a].right&&tagp[b].left)
                   {
                       x = i*m+j;
                       y = i*m+j+1;
                       Union(x,y);
                   }
               }
               if(i+1<n)//边界考虑
               {
                   if(tagp[a].down&&tagp[c].up)
                   {
                       x = i*m+j;
                       y = (i+1)*m+j;
                       Union(x,y);
                   }
               }
           }
       }
       int count = 0;
       int pos;
       for (i = 0; i < n; ++i)
       {
           for (j = 0; j < m; ++j)
           {
               pos = i*m+j;
               if(pos == find(pos))
               count++;
           }
       }
       printf("%d\n",count);
   }
   return 0;
}