UVa 127 "Accordian" Patience

UVa 127 "Accordian" Patience

不断地去寻找有没有可以移动的牌，直到不能移动。
以下是我的代码：

#include < iostream >
#include < string >
#include < stack >
#include < cstdio >
using   namespace  std;
const   int  kMaxn( 57 );

bool  Match( const   string   & a, const   string   & b)
{
     return  (a[ 0 ] == b[ 0 ]  ||  a[ 1 ] == b[ 1 ]);
}

stack < string >  s[kMaxn];

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

     int  n( 0 );
     string  t;
     while (cin >> t  &&  t != " # " )
    {
        s[n].push(t);
        n ++ ;
         if (n >= 52 )
        {
             while ( true )
            {
                 int  pos,sign;
                 for (pos = 0 ;pos < n;pos ++ )
                {
                     if (pos >= 3   &&  Match(s[pos].top(),s[pos - 3 ].top()))
                    {
                        sign = 1 ;
                         break ;
                    }
                     if (pos >= 1   &&  Match(s[pos].top(),s[pos - 1 ].top()))
                    {
                        sign = 2 ;
                         break ;
                    }
                }
                 if (pos == n)
                     break ;
                 if (sign == 1 )
                    s[pos - 3 ].push(s[pos].top());
                 else
                    s[pos - 1 ].push(s[pos].top());
                s[pos].pop();
                 if (s[pos].empty())
                {
                     for ( int  i = pos;i < n - 1 ;i ++ )
                        s[i] = s[i + 1 ];
                     while ( ! s[n - 1 ].empty())
                        s[n - 1 ].pop();
                    n -- ;
                }
            }

            cout << n;
            cout << "  pile " ;
             if (n > 1 )
                cout << " s " ;
            cout << "  remaining: " ;
             for ( int  i = 0 ;i < n;i ++ )
                cout << "   " << s[i].size();
            cout << endl;

             for ( int  i = 0 ;i < n;i ++ )
                 while ( ! s[i].empty())
                    s[i].pop();
            n = 0 ;
        }
    }

     return   0 ;
}