UVa 10651 Pebble Solitaire

UVa 10651 Pebble Solitaire

DP题，状态总量为2^12，其实可以用对称性减少一半状态数。
以下是我的代码：

/*
 * Author:  lee1r
 * Created Time:  2011/8/3 12:18:41
 * File Name: uva10651.cpp
  */
#include < iostream >
#include < sstream >
#include < fstream >
#include < vector >
#include < list >
#include < deque >
#include < queue >
#include < stack >
#include < map >
#include < set >
#include < bitset >
#include < algorithm >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cctype >
#include < cmath >
#include < ctime >
#define  L(x) ((x)<<1)
#define  R(x) (((x)<<1)+1)
#define  Half(x) ((x)>>1)
#define  lowbit(x) ((x)&(-(x)))
using   namespace  std;
const   int  kInf( 0x7f7f7f7f );
const   double  kEps(1e - 11 );
typedef  long   long  int64;
typedef unsigned  long   long  uint64;

int  d[ 4107 ];

int  dp( int  x)
{
     if (d[x] !=- 1 )
         return  d[x];
    d[x] = 0 ;
     for ( int  i = 0 ;i < 12 ;i ++ )
         if (x & ( 1 << i))
            d[x] ++ ;
     for ( int  i = 0 ;i < 10 ;i ++ )
         if ((x & ( 1 << i))  &&  (x & ( 1 << (i + 1 )))  &&   ! (x & ( 1 << (i + 2 ))))
        {
             int  newx(x);
            newx = ( ~ (( ~ newx) | ( 1 << i)));
            newx = ( ~ (( ~ newx) | ( 1 << (i + 1 ))));
            newx = (newx | ( 1 << (i + 2 )));
            d[x] = min(d[x],dp(newx));
        }
     for ( int  i = 0 ;i < 10 ;i ++ )
         if ( ! (x & ( 1 << i))  &&  (x & ( 1 << (i + 1 )))  &&  (x & ( 1 << (i + 2 ))))
        {
             int  newx(x);
            newx = ( ~ (( ~ newx) | ( 1 << (i + 1 ))));
            newx = ( ~ (( ~ newx) | ( 1 << (i + 2 ))));
            newx = (newx | ( 1 << i));
            d[x] = min(d[x],dp(newx));
        }
     return  d[x];
}

int  main()
{
     // freopen("data.in","r",stdin);
    
    memset(d, - 1 , sizeof (d));
    
     int  T;
    cin >> T;
     while (T -- )
    {
        string  s;
       cin >> s;
        int  n( 0 );
        for ( int  i = 0 ;i < s.size();i ++ )
            if (s[i] == ' o ' )
               n = (n | ( 1 << i));
       printf( " %d\n " ,dp(n));
    }
    
     return   0 ;
}