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http://poj.org/problem?id=3274
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
Output
Sample Input
7 3 7 6 7 2 1 4 2
Sample Output
4
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=100005;
const int hashseed=99983;
int n,k;
int sum[maxn][35],c[maxn][35];
int head[hashseed+2],next[maxn];
int hash(int v[])
{
int h=0;
for(int i=1; i<=k; i++)
{
h=(h<<2)+(v[i]>>4)^(v[i]<<10);
}
h=(h+hashseed)%hashseed;
return h;
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(sum,0,sizeof(sum));
for(int i=1; i<=n; i++)
{
int cnt;
scanf("%d",&cnt);
for(int j=1; j<=k; j++)
{
sum[i][j]=sum[i-1][j]+cnt%2;
cnt/=2;
}
}
int maxx=0;
memset(head,-1,sizeof(head));
for(int i=0; i<=n; i++)///一定要从0开始
{
for(int j=1; j<=k; j++)
{
c[i][j]=sum[i][j]-sum[i][1];
}
int h=hash(c[i]);
bool flag=0;
for(int p=head[h]; p!=-1; p=next[p])
{
int j;
for(j=1; j<=k; j++)
{
if(c[i][j]!=c[p][j])
break;
}
if(j>k)
{
maxx=max(maxx,i-p);
flag=1;
break;
}
}
if(flag==0)
{
next[i]=head[h];
head[h]=i;
}
}
printf("%d\n",maxx);
}
return 0;
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原文地址：http://blog.csdn.net/lvshubao1314/article/details/42218731
