# POJ 3261 Milk Patterns(后缀数组[可重叠的k次最长重复子串])

2016-10-27

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

# POJ 3261 Milk Patterns

Accept: 0 Submit: 0
Time Limit: 5000 MS Memory Limit : 65536 K

## Problem 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&#39;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 &le; N &le; 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 &le; K &le; 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

8 2
1
2
3
2
3
2
3
1

4

## Problem Idea

【题意】

8 2

1 2 3 2 3 2 3 1

【类型】

【分析】

"可重叠的k次最长重复子串"解法(摘自罗穗骞的国家集训队论文):

ps：只要有一个组里存在k个后缀,就说明当前二分情况有解

【时间复杂度&&优化】
O(nlogn)

## Source Code

```/*Sherlock and Watson and Adler*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<bitset>
#include<cmath>
#include<complex>
#include<string>
#include<algorithm>
#include<iostream>
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define bitnum(a) __builtin_popcount(a)
using namespace std;
const int N = 10;
const int M = 100005;
const int inf = 1000000007;
const int mod = 1000000007;
const int MAXN = 100010;
//rnk从0开始
//sa从1开始,因为最后一个字符(最小的)排在第0位
//height从1开始,因为表示的是sa[i - 1]和sa[i]
//倍增算法 O(nlogn)
int wa[MAXN], wb[MAXN], wv[MAXN], ws_[MAXN];
//Suffix函数的参数m代表字符串中字符的取值范围,是基数排序的一个参数,如果原序列都是字母可以直接取128,如果原序列本身都是整数的话,则m可以取比最大的整数大1的值
//待排序的字符串放在r数组中,从r[0]到r[n-1]，长度为n
//为了方便比较大小,可以在字符串后面添加一个字符,这个字符没有在前面的字符中出现过,而且比前面的字符都要小
//同上,为了函数操作的方便,约定除r[n-1]外所有的r[i]都大于0,r[n-1]=0
//函数结束后,结果放在sa数组中,从sa[0]到sa[n-1]
void Suffix(int *r, int *sa, int n, int m)
{
int i, j, k, *x = wa, *y = wb, *t;
//对长度为1的字符串排序
//一般来说,在字符串的题目中,r的最大值不会很大,所以这里使用了基数排序
//如果r的最大值很大,那么把这段代码改成快速排序
for(i = 0; i < m; ++i) ws_[i] = 0;
for(i = 0; i < n; ++i) ws_[x[i] = r[i]]++;//统计字符的个数
for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1];//统计不大于字符i的字符个数
for(i = n - 1; i >= 0; --i) sa[--ws_[x[i]]] = i;//计算字符排名
//基数排序
//x数组保存的值相当于是rank值
for(j = 1, k = 1; k < n; j *= 2, m = k)
{
//j是当前字符串的长度,数组y保存的是对第二关键字排序的结果
//第二关键字排序
for(k = 0, i = n - j; i < n; ++i) y[k++] = i;//第二关键字为0的排在前面
for(i = 0; i < n; ++i) if(sa[i] >= j) y[k++] = sa[i] - j;//长度为j的子串sa[i]应该是长度为2 * j的子串sa[i] - j的后缀（第二关键字）,对所有的长度为2 * j的子串根据第二关键字来排序
for(i = 0; i < n; ++i) wv[i] = x[y[i]];//提取第一关键字
//按第一关键字排序 (原理同对长度为1的字符串排序)
for(i = 0; i < m; ++i) ws_[i] = 0;
for(i = 0; i < n; ++i) ws_[wv[i]]++;
for(i = 1; i < m; ++i) ws_[i] += ws_[i - 1];
for(i = n - 1; i >= 0; --i) sa[--ws_[wv[i]]] = y[i];//按第一关键字,计算出了长度为2 * j的子串排名情况
//此时数组x是长度为j的子串的排名情况,数组y仍是根据第二关键字排序后的结果
//计算长度为2 * j的子串的排名情况,保存到数组x
t = x;
x = y;
y = t;
for(x[sa[0]] = 0, i = k = 1; i < n; ++i)
x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j]) ? k - 1 : k++;
//若长度为2 * j的子串sa[i]与sa[i - 1]完全相同,则他们有相同的排名
}
}
int Rank[MAXN], height[MAXN], sa[MAXN], r[MAXN];
void calheight(int *r,int *sa,int n)
{
int i,j,k=0;
for(i=1; i<=n; i++)Rank[sa[i]]=i;
for(i=0; i<n; height[Rank[i++]]=k)
for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++);
}
bool judge(int k,int c,int n)
{
int i,count=1;
for(i=1;i<n;i++)
{
if(height[i]>=c)
count++;
else
count=1;
if(count>=k)
return true;
}
return false;
}
int main()
{
int n,k,i,Max,L,R,mid,ans;
while(~scanf("%d%d",&n,&k))
{
Max=0;
for(i=0;i<n;i++)
{
scanf("%d",&r[i]);
Max=max(Max,r[i]);
}
r[i]=0;
Suffix(r,sa,n+1,Max+1);
calheight(r,sa,n);
L=0,R=n,ans=0;
while(L<=R)
{
mid=(L+R)/2;
if(judge(k,mid,n+1))
L=mid+1,ans=max(ans,mid);
else
R=mid-1;
}
printf("%d\n",ans);
}
return 0;
}```