ZOJ 2412 Farm Irrigation

ZOJ 2412 Farm Irrigation

题目大意：给出水管的分布，求出需要多少个水源。即求出无向图有多少个连通分量。
DFS遍历一次即可。
以下是我的代码：

#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxn( 57 );
const   int  kWater[] = { 6 , 10 , 20 , 24 , 18 , 12 , 14 , 22 , 28 , 26 , 30 };
const   int  dx[] = { 0 , 0 , 1 , - 1 },dy[] = { 1 , - 1 , 0 , 0 };

int  m,n;
char  g[kMaxn][kMaxn];
bool  used[kMaxn][kMaxn];

void  dfs( int  x, int  y)
{
    used[x][y] = true ;
     for ( int  i = 0 ;i < 4 ;i ++ )
    {
         int  xx(x + dx[i]),yy(y + dy[i]);
         if (xx < 1   ||  xx > m  ||  yy < 1   ||  yy > n  ||  used[xx][yy])
             continue ;

         int  t1(g[x][y] - ' A ' ),t2(g[xx][yy] - ' A ' );
         if (i == 0   &&  (kWater[t1] & ( 1 << 3 ))  &&  (kWater[t2] & ( 1 << 2 )))
            dfs(xx,yy);
         else   if (i == 1   &&  (kWater[t1] & ( 1 << 2 ))  &&  (kWater[t2] & ( 1 << 3 )))
            dfs(xx,yy);
         else   if (i == 2   &&  (kWater[t1] & ( 1 << 4 ))  &&  (kWater[t2] & ( 1 << 1 )))
            dfs(xx,yy);
         else   if (i == 3   &&  (kWater[t1] & ( 1 << 1 ))  &&  (kWater[t2] & ( 1 << 4 )))
            dfs(xx,yy);
    }
}

int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */

     while (scanf( " %d%d " , & m, & n) == 2   &&  m >= 0   &&  n >= 0 )
    {
         for ( int  i = 1 ;i <= m;i ++ )
        {
            getchar();
             for ( int  j = 1 ;j <= n;j ++ )
                scanf( " %c " , & g[i][j]);
        }

         int  ans( 0 );
        memset(used, false ,kMaxn * kMaxn * sizeof ( bool ));
         for ( int  i = 1 ;i <= m;i ++ )
             for ( int  j = 1 ;j <= n;j ++ )
                 if ( ! used[i][j])
                {
                    ans ++ ;
                    dfs(i,j);
                }

        printf( " %d\n " ,ans);
    }

     return   0 ;
}