POJ 3261 后缀数组分组二分

Milk Patterns

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 9121 Accepted: 4095Case 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 overN (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 leastK times.

Input

Line 1: Two space-separated integers: N andK
Lines 2..N+1: N integers, one per line, the quality of the milk on dayi appears on theith line.

Output

Line 1: One integer, the length of the longest pattern which occurs at leastK times

Sample Input

8 212323231

Sample Output

4

题意：求至少重复k次的最长子串，子串之间可以重叠。

解题思路：后缀数组的简单应用，求出sa，rank，height，然后二分长度分组，判断可行性。

代码：

/* ***********************************************Author :xianxingwuguanCreated Time :2014-1-27 17:41:59File Name :1.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const int maxn=222222;int height[maxn],rank[maxn],sa[maxn],c[1000100],t1[maxn],t2[maxn],str[maxn];void da(int *str,int n,int m){      int i,j,k,p,*x=t1,*y=t2;      for(i=0;i<m;i++)c[i]=0;      for(i=0;i<n;i++)c[x[i]=str[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(k=1;k<=n;k<<=1)      {     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-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;     if(p>=n)break;\     m=p;      }}void calheight(int *str,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(str[i+k]==str[j+k])k++;     height[rank[i]]=k;      }}bool judge(int *sa,int n,int m,int k){      int s=1;      for(int i=1;i<=n;i++)      {     if(height[i]>=m)     {    s++;    if(s>=k)return 1;     }     else s=1;      }      return 0;}int main(){      //freopen("data.in","r",stdin);      //freopen("data.out","w",stdout);      int n,i,j,k,m;      while(~scanf("%d%d",&n,&k))      {     for(i=0;i<n;i++)scanf("%d",&str[i]);     str[n]=0;     da(str,n+1,1000100);     calheight(str,n);     int l=1,r=n;     while(l<r)     {    m=(l+r+1)>>1;    if(judge(sa,n,m,k))l=m;    else r=m-1;     }     printf("%d\n",l);      }      return 0;}