Poj 3261 Milk Patterns(后缀数组+二分答案)

Milk Patterns

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

Source

USACO 2006 December Gold

/*
后缀数组+二分答案.
求可重叠k次的最长重复子串.
比较暴力的做法,并没有离散化.
随便搞搞就行了..
*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define MAXN 1000010
using namespace std;
int n,K,m=1000002,ans,sa[MAXN],rank1[MAXN],c[MAXN],ht[MAXN],t1[MAXN],t2[MAXN],s[MAXN];
bool cmp(int *y,int a,int b,int k)
{
    int a1=y[a],b1=y[b];
    int a2=a+k>=n?-1:y[a+k];
    int b2=b+k>=n?-1:y[b+k];
    return a1==b1&&a2==b2;
}
void slovesa()
{
    int *x=t1,*y=t2;
    for(int i=0;i<m;i++) c[i]=0;
    for(int i=0;i<n;i++) c[x[i]=s[i]]++;
    for(int i=1;i<m;i++) c[i]+=c[i-1];
    for(int i=n-1;i>=0;i--) sa[--c[x[i]]]=i;
    for(int k=1,p=0;k<=n;k<<=1,m=p,p=0)
    {
        for(int i=n-k;i<n;i++) y[p++]=i;
        for(int i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(int i=0;i<m;i++) c[i]=0;
        for(int i=0;i<n;i++) c[x[y[i]]]++;
        for(int i=1;i<m;i++) c[i]+=c[i-1];
        for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
        swap(x,y),p=1,x[sa[0]]=0;
        for(int i=0;i<n;i++)
        {
            if(cmp(y,sa[i-1],sa[i],k)) x[sa[i]]=p-1;
            else x[sa[i]]=p++;
        }
        if(p>=n) break;
    }
}
void sloveheight()
{
    int k=0;
    for(int i=0;i<n;i++) rank1[sa[i]]=i;
    for(int i=0;i<n;ht[rank1[i++]]=k)
    {
        int j=sa[rank1[i]-1];
        if(k) k--;
        while(j+k<n&&i+k<n&&s[i+k]==s[j+k]) k++;
    }
    ht[0]=0;
}
bool check(int x)
{
    int tot=1;bool flag=false;
    for(int i=1;i<n;i++)
    {
        if(ht[i]>=x) tot++;
        else tot=1;
        if(tot>=K) return true;
    }
    return false;
}
void erfen(int l,int r)
{
    int mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(check(mid)) ans=mid,l=mid+1;
        else r=mid-1;
    }
}
int main()
{
    scanf("%d%d",&n,&K);
    for(int i=0;i<n;i++) scanf("%d",&s[i]),s[i]++;
    s[++n]=0;
    slovesa(),sloveheight(),erfen(0,n);
    printf("%d\n",ans);
    return 0;
}