POJ 3261 Milk Patterns

Description

Farmer John has noticed that the qualityof milk given by his cows varies from day to day. On further investigation, hediscovered that although he can't predict the quality of milk from one day tothe next, there are some regular patterns in the daily milk quality.

To perform a rigorous study, he hasinvented a complex classification scheme by which each milk sample is recordedas an integer between 0 and 1,000,000 inclusive, and has recorded data from asingle cow over N (1 ≤N ≤ 20,000) days. Hewishes to find the longest pattern of samples which repeats identically atleast K (2 ≤ K ≤ N) times. This mayinclude overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, forexample.

Help Farmer John by finding the longestrepeating subsequence in the sequence of samples. It is guaranteed that atleast one subsequence is repeated at least K times.

Input

Line 1: Twospace-separated integers: N and K 
Lines 2..N+1: N integers, one perline, the quality of the milk on day i appears on the ith line.

Output

Line 1: Oneinteger, the length of the longest pattern which occurs at least K times

Sample Input

8 2

1

2

3

2

3

2

3

1

Sample Output

4

题目大意：求一个含有N个数字的串中，至少出现了K次的子串的最大长度。

 

先用后缀数组求出height值，接着问题就可以转化成，height数组中连续k-1个中的最小值的最大值。

 

#include
#include
#include
#include
#include
#include
#define MAX 1000010
using namespace std;

int n ,k ,num[MAX];
int wa[MAX] ,wb[MAX] ,wv[MAX] ,ws1[MAX] ,sa[MAX];
int rank1[MAX],height[MAX];

int cmp(int *r,int a,int b,int l)
{
	return r[a]==r[b]&&r[a+l]==r[b+l];
}

void da(int *r,int *sa,int n,int m)
{
	int i,j,p,*x=wa,*y=wb,*t;
	for(i=0;i=0;i--) sa[--ws1[x[i]]]=i;
	for(j=1,p=1;p=j) y[p++]=sa[i]-j;
		for(i=0;i=0;i--) sa[--ws1[wv[i]]]=y[i];
		for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i q;
		int ans = 0;
		for(int i = 1;i <= n;i++)
		{
			while(!q.empty()&&height[i] <= height[q.back()])
			{
				q.pop_back();
			}
			while(!q.empty()&&q.front() <= i - k + 1)
			{
				q.pop_front();
			}
			q.push_back(i);
			if(i + 1 >= k && height[q.front()] > ans)
			{
				ans = height[q.front()];
			}
		}

		printf("%d\n",ans);
	}
	return 0;
}