POJ 3274, Gold Balanced Lineup

Hash Table

hash function

1.

for(int j=0; j<K; ++j)
   key=((key<<2)+(cnt[j]>>4))^(cnt[j]<<10);

2.

z = cnt.size()

for(int j=0; j < K; ++j)

   z = 3179（质数）* z + cnt[j];

Description

Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might have spots, cows exhibiting feature #2 might prefer C to Pascal, and so on.

FJ has even devised a concise way to describe each cow in terms of its "feature ID", a single K-bit integer whose binary representation tells us the set of features exhibited by the cow. As an example, suppose a cow has feature ID = 13. Since 13 written in binary is 1101, this means our cow exhibits features 1, 3, and 4 (reading right to left), but not feature 2. More generally, we find a 1 in the 2^(i-1) place if a cow exhibits feature i.

Always the sensitive fellow, FJ lined up cows 1..N in a long row and noticed that certain ranges of cows are somewhat "balanced" in terms of the features the exhibit. A contiguous range of cows i..j is balanced if each of the K possible features is exhibited by the same number of cows in the range. FJ is curious as to the size of the largest balanced range of cows. See if you can determine it.

 

Input
Line 1: Two space-separated integers, N and K.
Lines 2..N+1: Line i+1 contains a single K-bit integer specifying the features present in cow i. The least-significant bit of this integer is 1 if the cow exhibits feature #1, and the most-significant bit is 1 if the cow exhibits feature #K.

 

Output
Line 1: A single integer giving the size of the largest contiguous balanced group of cows.

 

Sample Input
7 3
7
6
7
2
1
4
2

 

Sample Output
4

Hint
In the range from cow #3 to cow #6 (of size 4), each feature appears in exactly 2 cows in this range

 

Source
USACO 2007 March Gold

---

//  POJ3274.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < algorithm >
using   namespace  std;

struct  Cow
{
    Cow():next( 0 ),pos( - 1 ){}
    Cow *  next;
     int  step[ 30 ];
     int  pos;
};
bool  compare( int  c1[],  int  c2[],  int  K)
{
     for  ( int  i  =   0 ; i  <  K;  ++ i)
         if  (c1[i]  !=  c2[i]) return   false ;
     return   true ;
};
inline  int  calcKey( int  cnt[],  int  K,  int  SIZE)
{
     int  key  =   0 ;
     for ( int  j = 0 ; j < K;  ++ j)
        key = ((key << 2 ) + (cnt[j] >> 4 )) ^ (cnt[j] << 10 );
    key  =  key  %  SIZE;
    key  =  key  <   0   ?  key  +  SIZE : key;
     return  key;
}
int  main( int  argc,  char *  argv[])
{
     int  cnt[ 30 ];
     const   int  SIZE  =   33119 ;
    memset(cnt,  0 ,  sizeof (cnt));
    Cow hash[SIZE];
    memset(hash,  0 ,  sizeof (hash));
     int  mb  =   0 ;
    
     int  N, K;
    scanf( " %d %d\n " ,  & N,  & K);

     // init hash table
    Cow *  c  =   new  Cow;
    c -> pos  =   0 ;
    copy( & cnt[ 0 ],  & cnt[K],  & c -> step[ 0 ]);
    hash[ 0 ].next  =  c;

     int  feature;
     for  ( int  i  =   1 ; i  <=  N;  ++ i)
    {
        scanf( " %d " ,  & feature);

         // update cnt[K]
         for  ( int  j  =   0 ; j  <  K;  ++ j)
             if  (((feature >> j)  &   1 )  ==   1 )  ++ cnt[j];
         if  ( * min_element( & cnt[ 0 ],  & cnt[K])  !=   0 )
             for  ( int  j  =   0 ; j  <  K;  ++ j)  -- cnt[j];

         int  key  =  calcKey(cnt,K,SIZE);
         // search key
        Cow *  pt  =   & hash[key];
         bool  found  =   false ;
         while  (pt -> next  !=  NULL)
        {
            pt  =  pt -> next;
             if  (compare(pt -> step,cnt, K))
            {
                mb  =  max(i  -  pt -> pos, mb);
                found  =   true ;
                 break ;
            }
        }
        
         // update hash table
         if  (found  ==   false )
        {
            Cow *  pc  =   new  Cow;
            pc -> pos  =  i;
            copy( & cnt[ 0 ],  & cnt[K],  & pc -> step[ 0 ]);
            pt -> next  =  pc;
        }
    }
    cout  <<  mb  <<  endl;
     return   0 ;
}