POJ 3261 Milk Patterns (求可重叠的k次最长重复子串)

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 14094		Accepted: 6244
Case Time Limit: 2000MS

Description

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

To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

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

Input

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

Output

Line 1: One integer, 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

/*
 * POJ 3261 Milk Patterns
 * 求可重叠的k次最长重复子串
 *
 * 二分答案，每次直接到height数组里面check即可。
 */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 20000+100;

int sa[MAXN];
int t1[MAXN],t2[MAXN],c[MAXN];
int rank[MAXN],height[MAXN];
void build_sa(int s[],int n,int m)
{
    int i,j,p,*x=t1,*y=t2;
    for(i=0;i<m;i++)c[i]=0;
    for(i=0;i<n;i++)c[x[i]=s[i]]++;
    for(i=1;i<m;i++)c[i]+=c[i-1];
    for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;
    for(j=1;j<=n;j<<=1)
    {
        p=0;
        for(i=n-j;i<n;i++)y[p++]=i;
        for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
        for(i=0;i<m;i++)c[i]=0;
        for(i=0;i<n;i++)c[x[y[i]]]++;
        for(i=1;i<m;i++)c[i]+=c[i-1];
        for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];
        swap(x,y);
        p=1;x[sa[0]]=0;
        for(i=1;i<n;i++)
            x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
        if(p>=n)break;
        m=p;
    }
}
void getHeight(int s[],int n)
{
    int i,j,k=0;
    for(i=0;i<=n;i++)rank[sa[i]]=i;
    for(i=0;i<n;i++)
    {
        if(k)k--;
        j=sa[rank[i]-1];
        while(s[i+k]==s[j+k])k++;
        height[rank[i]]=k;
    }
}
int ss[MAXN];
bool check(int n,int k,int len)
{
    int cnt=1;
    for(int i=2;i<=n;i++)
    {
        if(height[i]>=len)
        {
            cnt++;
            if(cnt>=k) return true;
        }
        else cnt=1;
    }
    return false;
}

int main()
{
    int n,k;
    int ma=0;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&ss[i]);
        ma=max(ma,ss[i]);
    }
    ss[n]=0;
    build_sa(ss,n+1,ma+1);
    getHeight(ss,n);
    int l=0,r=n,ans;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(check(n,k,mid)) l=mid+1,ans=mid;
        else r=mid-1;
    }
    printf("%d
",ans);
    return 0;
}