Milk Patterns poj 3261 后缀数组 - 可重叠的K次最长重复子串

poj 3261

题目连接： 点击打开链接

 

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 Ktimes.

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

 

            题意：找出出现k次的可重叠的最长子串的长度

                求可重叠的k次最长重复子串

                分析：

                    先二分答案，然后将后缀分成若干组。不同的是，这里要判断的是有没有一个组的后缀个数不小于

                    K，如果有，那么存在k个相同的子串满足条件，否则不存在，算法时间复杂度：O(nlongn)

        

/*
@Author: Top_Spirit
@Language: C++
*/
//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std ;
typedef unsigned long long  ull ;
typedef long long ll ;
//const int Maxn = 1e5 + 10 ;
const int INF = 0x3f3f3f3f ;
const ull seed = 233 ;
const ull MOD = 1e4 + 7 ;
const int maxn = 2e4 + 10;
const int Max = 10000;

int sa[maxn],Rank[maxn],height[maxn];
int wa[maxn],wb[maxn],wv[maxn],Ws[maxn];
int num[maxn],s[maxn];

int cmp (int *r,int a,int b,int l){
    return r[a] == r[b] && r[a + l] == r[b + l];
}

void get_sa (int * r, int n, int m){
    int i, j , p, *x = wa,*y = wb,*t;
    for (i = 0; i < m; i++) Ws[i] = 0;
    for (i = 0; i < n; i++) Ws[x[i] = r[i]]++;
    for (i = 1; i < m; i++) Ws[i] += Ws[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--Ws[x[i]]] = i;
    for (j = 1, p = 1 ; p < n; j *= 2, m = p){
        for (p = 0, 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 < n; i++) wv[i] = x[y[i]];
        for (i = 0; i < m; i++) Ws[i] = 0;
        for (i = 0; i < n; i++) Ws[wv[i]]++;
        for (i = 0; i < m; i++) Ws[i] += Ws[i - 1];
        for (i = n - 1; i >= 0; i--) sa[--Ws[wv[i]]] = y[i];
        for (t = x, x = y, y = t,p = 1, x[sa[0]] = 0, i = 1; i < n; i++){
            x[sa[i]] = cmp(y,sa[i - 1], sa[i], j) ? p -1 : p++;
        }
    }
}
void get_height(int *r,int n){
    int k = 0, j;
    for (int i = 1; i <= n; i++) Rank[sa[i]] = i;
    for (int 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 f (int n, int k, int x){
    for (int i = 2; i <= n; i++){
        if (height[i] >= x){
            int cmt = 1;
            for (i; i <= n && height[i] >= x; i++) cmt++;
            if (cmt >= k) return 1;
        }
    }
    return 0;
}

int solve (int n, int k){
    int ans = 0;
    int ls = 0, rs = n;
    while (ls <= rs){
        int mid = (ls + rs) / 2;
        if (f(n,k,mid)) ans = mid, ls = mid + 1;
        else rs = mid - 1;
    }
    return ans ;
}
int main (){
    int n , k ;
    scanf("%d%d",&n,&k);
    for (int i = 0 ;i < n; i++){
        scanf("%d",&num[i]);
    }
    num[n] = 0;
    get_sa(num,n+1,Max);
    get_height(num,n);
//    for (int i = 0; i < n; i++){
//        printf("height - > %d ",height[i]);
//    }
    printf("%d\n",solve(n,k));
}