ACM: hash题 poj 3274 (题目看了好…

                                                        Gold Balanced Lineup

Description

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

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

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

Input

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

Output

Line 1: A single integer giving the size of the largestcontiguous 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 featureappears in exactly 2 cows in this range

 

题意: 现在有N头奶牛, 每头奶牛有一个特征值,这个数是由K位二进制表示,对应的位数为1的代表具有该’位‘特征

        牛排成一列,求最长的连续区间.(这里理解好久的最长区间).

解题思路:

               1.假设sum[i][j]表示前i头奶牛的第j特征的和. 在区间[a+1,b]之间.

               有: sum[b][1] - sum[a][1] = sum[b][2] - sum[b][1] = sum[b][j] -sum[a][j]  (1<=j<=K )

               2. 公式变形:

                   sum[b][1] - sum[b][2] = sum[a][1] - sum[b][1]

                   sum[b][1] - sum[b][j] = sum[a][1] - sum[a][j]

               即: sum[i][1] - sum[i][2] 也是个常量.

                设: c[i][j] = sum[i][1] -sum[i][2];  (1<=j<=K )

               3. 在网上查找讨论发现对c[i]数组进行hash查找. 先将数组转化字符串用UNIX哈希函数.

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 100003
#define KK 33
#define MAXHASH 10007

int a[MAX];
int sum[MAX][KK];
int c[MAX][KK];
int n, k;
int head[MAXHASH+5];
int next[MAX];
int ans;

int elfHash(char *key)
{
    unsignedlong h = 0;
   while(*key)
    {
      h = (h << 4) +*key++;
      unsigned long g = h &0xf0000000L;
      if(g) h ^= g >>24;
      h &= ~g;
    }
    return(h+MAXHASH) % MAXHASH;
}

int getHash(int num[])
{
    charstr[KK];
    int i;
    for(i = 1; i<= k; ++i)
      str[i] = num[i]+'a';
    str[i] ='\0';
    returnelfHash(str+1);
}

void run()
{
   memset(head,-1,sizeof(head));
    for(int i =0; i <= n; ++i)
    {
      for(int j = 1; j <= k; ++j)
         c[i][j] = sum[i][j] - sum[i][1];
      int h = getHash(c[i]);
      int flag = 0;
      for(int e = head[h]; e != -1; e = next[e])
      {
         int t;
         for(t = 1; t <= k; ++t)
         {
            if(c[i][t] != c[e][t])
               break;
         }
         if(t > k)
         {
            if(i-e > ans)
               ans = i-e;
            flag = 1;
            break;
         }
      }
      if(flag == 0)
      {
         next[i] = head[h];
         head[h] = i;
      }
    }
   printf("%d\n",ans);
}

int main()
{
//   freopen("input.txt","r",stdin);
   while(scanf("%d%d",&n,&k) != EOF)
    {
      for(int i = 1; i <= n; ++i)
         scanf("%d",&a[i]);
         
      memset(sum,0,sizeof(sum));      
      for(int i = 1; i <= n; ++i)
      {
         int t = a[i];
         for(int j = 1; j <= k; ++j)
         {
            sum[i][j] = sum[i-1][j] + t%2;
            t /= 2;
         }
      }
      
      ans = 0;
      run();
    }
    return0;
}