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

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 16742		Accepted: 7390
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

/*************************************************************************
    > File Name: poj3261.cpp
    > Author: LyuCheng
    > Created Time: 2017-11-30 17:04
    > Description: 询问最长可重复k次的串的长度，求出height数组的值，求连续
            的lca是不是大于k如果大于k那么就满足条件
 ************************************************************************/
#include <bits/stdc++.h>
#include <stdio.h>
#include <string.h>

#define maxn 23456

using namespace std;

struct SuffixArray{
    int s[maxn];
    int sa[maxn];
    int rank[maxn];
    int height[maxn];
    int t1[maxn],t2[maxn],c[maxn],n;
    void build_sa(int m){
        int i,*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(int k=1;k<=n;k<<=1){
            int p=0;
            for(i=n-k;i<n;i++)y[p++]=i;
            for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
            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]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]?p-1:p++;
            if(p>=n) break;
            m=p;
        }
    }
    void build_height(){
        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;
        }
    }
}sa;

int n,k;
int maxs;
int ans;

inline bool ok(int pos){
    int s=1;
    for(int i=1;i<n;i++){
        if(sa.height[i]>=pos){
            s++;
            if(s>=k)
                return true;
        }else{
            s=1;
        }
    }
    return false;
}

inline void init(){
    maxs=-1;
    ans=0;
}

int main(){
    while(scanf("%d%d",&n,&k)!=EOF){
        init();
        sa.n=n;
        for(int i=0;i<n;i++){
            scanf("%d",&sa.s[i]);
            maxs=max(maxs,sa.s[i]);
        }
        sa.s[n]=1000001;
        sa.build_sa(maxs+1); 
        sa.build_height();
        int l=0,r=n,m;
        while(l<=r){
            m=(l+r)>>1;
            if(ok(m)==true){
                ans=m;
                l=m+1;
            }else{
                r=m-1;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}