POJ 3274 Gold Balanced Lineup（Hash）

Gold Balanced Lineup

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7983		Accepted: 2357

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 onlyK 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

 解题报告：这道题看了好长时间还是没有看懂，没有想到这个题为什么和Hash建立关系， 就在网上找了一位大牛的解法，下面是摘的大牛的：

数组sum[i][j]表示从的1到i头cow属性j的和。所以题目要求等价为求满足

sum[i][0]-sum[j][0]==sum[i][1]-sum[j][1]==.....==sum[i][k-1]-sum[j][k-1] (j<i)

最大的i-j

将上式变换为

sum[i][1]-sum[i][0]==sum[j][1]-sum[j][0]

sum[i][2]-sum[i][0]==sum[j][2]-sum[j][0]

.

.

sum[i][k-1]-sum[i][0]==sum[j][k-1]-sum[j][0]

令C[i][l]=sum[i][l]-sum[i][0] (0<l<k)。

所以只需求满足C[i]==C[j] 中最大的i-j。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX = 100007;
int hash[MAX], sum[MAX][35], binary[MAX][35], c[MAX][35];
int SearchHash(int *v, int n)//这是网上淘来的，自己写的hash提交时就错！
{
       int i, p=0;
       for(i=0; i<n; i++)
       {
              p=((p<<2)+(v[i]>>4))^(v[i]<<10);
       }
       p = p % MAX;
       if (p <0 )   p = p + MAX;
       return p;
}
int main()
{
    int N, K, i, j, a, p;
    while (scanf("%d%d", &N, &K) != EOF)
    {
        memset(hash, -1, sizeof(hash));
        memset(sum, 0, sizeof(sum));
        memset(binary, 0, sizeof(binary));
        memset(c, 0, sizeof(c));
        int ans = 0;
        hash[SearchHash(c[0], K)] = 0;
        for (i = 1; i <= N; ++i)
        {
            scanf("%d", &a);
            for (j = 0; j < K; ++j)
            {
                binary[i][j] = a & 1;//相当于binary[i][j] = a % 2;
                a = a >> 1;//相当于a = a / 2;
                sum[i][j] = sum[i - 1][j] + binary[i][j];//求各自的和
                c[i][j] = sum[i][j] - sum[i][0];
            }
            p = SearchHash(c[i], K);
            if (hash[p] != -1)
            {
                for (j = 1; j < K; ++j)
                {
                    if (c[i][j] != c[hash[p]][j])
                    {
                        break;
                    }
                }
                if (j == K)
                {
                    if (i - hash[p] > ans)
                    {
                        ans = i - hash[p];
                    }
                }
            }
            if (hash[p] == -1)
            {
                hash[p] = i;
            }
        }
        printf("%d\n", ans);
    }
}