poj 3274 Gold Balanced Lineup

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本文介绍了一种算法，用于解决 Farmer John 的牛群中找到最大的连续平衡子集的问题。这些牛具有不同的特征，每头牛由一个 K 位的二进制数表示其拥有的特征。平衡子集定义为在该范围内每一种特征出现的次数相同。

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

Sample Output

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 100010
#define Mod 1001007
using namespace std;

int cow[MAX][40];
int d[MAX][40];
int head[Mod+100];
int n,k;

int Hash(int s[])
{
    int p = 0;
    for(int i=0; i<k; i++)
        p = ((p<<2)+(s[i]>>4))^(s[i]<<10);
    p = p % Mod;
    p = p < 0 ? p + Mod : p;
    return p;
}
int main()
{
    std::ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    while(cin>>n>>k)
    {
        int ans = 0;
        memset(cow,0,sizeof(cow));
        memset(d,0,sizeof(d));
        memset(head,-1,sizeof(head));
        head[Hash(d[0])] = 0;
        for(int i=1; i<=n; i++)
        {
            int x;
            cin>>x;
            for(int j=0; j<k; j++)
            {
                cow[i][j] = x & 1;
                x >>= 1;
                cow[i][j] += cow[i-1][j];
                d[i][j] = cow[i][j] - cow[i][0];
            }

            int h = Hash(d[i]);
            while(head[h] != -1)
            {
                if(memcmp(d[head[h]],d[i],sizeof(d[i])) == 0)
                {
                    if(ans < i - head[h])
                    {
                        ans = i - head[h];
                        break;
                    }
                }
                h++;
            }
            if(head[h] == -1)
                head[h] = i;
        }
        cout<<ans<<endl;
    }
    return 0;
}

另一个版本

#include <iostream>
#include <cstdio>
#include <cstring>

#define MAX 100010
#define Mod 100007
using namespace std;

int cow[MAX][33];
int head[Mod],next[Mod];
int n,k;

int Hash(int s[])
{
    int p = 0;
    for(int i=0;i<k;i++)
        p = ((p<<2)+(s[i]>>4))^(s[i]<<10);
    p = p % Mod;
    p = p < 0 ? p + Mod : p;
    return p;
}
int main()
{
    std::ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    while(cin>>n>>k)
    {
        memset(cow,0,sizeof(cow));
        memset(head,-1,sizeof(head));
        memset(next,0,sizeof(next));
        for(int i=1;i<=n;i++)
        {
            int x;
            cin>>x;
            for(int j=0;j<k;j++)
            {
                cow[i][j] = x & 1;
                x >>= 1;
                if(i > 1)
                    cow[i][j] += cow[i-1][j];
            }
        }

        int ans = 0;
        for(int i=0;i<=n;i++)
        {
            int temp = cow[i][0];
            for(int j=0;j<k;j++)
                cow[i][j] -= temp;
            int h = Hash(cow[i]);
            bool flag = false;
            for(int j = head[h]; j!=-1; j = next[j])
            {
                if(memcmp(cow[j],cow[i],sizeof(cow[i])) == 0)
                {
                    flag = true;
                    ans = max(ans,i-j);
                    break;
                }
            }
            if(!flag)
            {
                next[i] = head[h];
                head[h] = i;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}