UVa 11205 The broken pedometer

UVa 11205 The broken pedometer

枚举含p个元素的集合的子集，求出该子集和给定的若干个集合的交集，如果得到的这些交集都不相同，则该子集符合条件。题目要求求出符合条件的子集中最少含有多少个元素。
以下是我的代码：

#include < bitset >
#include < algorithm >
#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxp( 17 );
const   int  kMaxn( 107 );

bool  cmp( const  bitset < kMaxp >   & a, const  bitset < kMaxp >   & b)
{
     return  (a.to_ulong() < b.to_ulong());
}

int  p,n,ans;
bitset < kMaxp >  r[kMaxn];
bitset < kMaxp >  used,t[kMaxn];

bool  OK()
{
     for ( int  i = 1 ;i < n;i ++ )
         if (t[i] == t[i + 1 ])
             return   false ;
     return   true ;
}

void  dfs( int  depth)
{
     if (depth > p)
    {
         for ( int  i = 1 ;i <= n;i ++ )
            t[i] = r[i];
         for ( int  i = 1 ;i <= n;i ++ )
            t[i] = t[i] & used;
        sort(t + 1 ,t + n + 1 ,cmp);
         if (OK())
            ans = min(ans,( int )used.count());
         return ;
    }
    dfs(depth + 1 );
    used[depth] = 1 ;
    dfs(depth + 1 );
    used[depth] = 0 ;
}

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

     int  T;
    scanf( " %d " , & T);
     while (T -- )
    {
        scanf( " %d%d " , & p, & n);
         for ( int  i = 1 ;i <= n;i ++ )
            r[i].reset();
         for ( int  i = 1 ;i <= n;i ++ )
             for ( int  j = 1 ;j <= p;j ++ )
            {
                 int  t;
                scanf( " %d " , & t);
                r[i][j] = t;
            }

        ans = 315 ;
        used.reset();
        dfs( 1 );

        printf( " %d\n " ,ans);
    }

     return   0 ;
}