三部曲一(数据结构)-1022-Gold Balanced Lineup

Gold Balanced Lineup

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 3

Problem 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 Kdifferent 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

 

Source

PKU

 

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <cmath>
#include <stdlib.h>
#define maxn 100010
#define seed 999983
using namespace std;

int n,k;
int feature[maxn][31],sum[maxn][31],c[maxn][31],maxlen=0;

struct Hash
{
    int id,next;
} h[1000000];

bool compair(int a,int b)
{
    int i;
    for(i=0; i<k; i++)
    {
        if(c[a][i]!=c[b][i])
            return false;
    }
    return true;
}

int count_key(int index)
{
    int i,key=0;
    //Hash公式
    for(i=0; i<k; i++)
        key=((key<<2)+(c[index][i]>>4))^(c[index][i]<<10);
    key=key%seed;
    if(key<0)
        key=key+seed;
    return key;
}

//int count_key(int index)
//{
//    int i,key=0;
//    for(i=0; i<k; i++)
//    {
//        key+=abs(c[index][i]);
//        key%=seed;
//    }
//    return key;
//}

void InsertAndCheck(int index)
{
    int i,key,last=0;
    key=count_key(index);
//    cout<<"key="<<key<<endl;
    //    cout<<key<<endl;
    if(h[key].id==-1)
    {
        h[key].id=index;
    }
    else
    {
        bool found=false;
        for(i=key; i!=-1; i=h[i].next)
        {
//            cout<<"run "<<index<<' '<<i<<endl;
            last=i;
            if(!found&&compair(index,h[i].id))
            {
                if(index-h[i].id>maxlen)
                {
                    maxlen=abs(h[i].id-index);
                    found=true;
                }
            }
        }
        for(i=last; h[i].id!=-1; i=(i+1)%seed);
        h[last].next=i;
        h[i].id=index;
    }
}

int main()
{
//    freopen("in.txt","r",stdin);
    memset(h,-1,sizeof(h));
    scanf("%d%d",&n,&k);
    int i,j,tmp;
    n++;
    for(i=0; i<maxn; i++)
        h[i].next=-1;
    InsertAndCheck(1);
    for(i=2; i<=n; i++)
    {
        scanf("%d",&tmp);
        for(j=0; j<k; j++)
        {
            feature[i][j]=tmp%2;
            sum[i][j]=sum[i-1][j]+feature[i][j];
            c[i][j]=sum[i][j]-sum[i][0];
            tmp>>=1;
        }
        InsertAndCheck(i);
    }
////    for(i=0;i<=n;i++)
////    {
////        for(j=0;j<k;j++)
////            cout<<c[i][j]<<' ';
////        cout<<endl;
////    }
//    for(i=0;i<6;i++)
//        cout<<h[i].id<<' '<<h[i].next<<endl;
    printf("%d
",maxlen);
    return 0;
}