poj3261 后缀数组求重复k次可重叠的子串的最长长度

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 13669		Accepted: 6041
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

 

题意:

n个数，求重复k次的，可重叠子串的最长长度。

思路:

求出height数组，然后二分答案，再判断连续的height[i] >= mid的个数是否大于k。

 

/*
 * Author:  sweat123
 * Created Time:  2016/6/28 16:24:57
 * File Name: main.cpp
 */
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 20010;
int wa[MAXN],wb[MAXN],wc[1000010],n,height[MAXN],sa[MAXN],k,r[MAXN],Rank[MAXN];
void da(int *r,int *sa,int n,int m){
    int *x = wa,*y = wb;
    for(int i = 0; i < m; i++)wc[i] = 0;
    for(int i = 0; i < n; i++)wc[x[i] = r[i]] ++;
    for(int i = 0; i < m; i++)wc[i] += wc[i-1];
    for(int i = n - 1; i >= 0; i--)sa[--wc[x[i]]] = i;
    for(int p = 1,k = 1; p < n; m = p,k <<= 1){
        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++)wc[i] = 0;
        for(int i = 0; i < n; i++)wc[x[y[i]]] ++;
        for(int i = 0; i < m; i++)wc[i] += wc[i-1];
        for(int i = n - 1; i >= 0; i--)sa[--wc[x[y[i]]]] = y[i];
        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for(int 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++;
    }   
}
void calheight(int *r,int *sa,int n){
    for(int i = 1; i <= n; i++)Rank[sa[i]] = i;
    int k = 0,j;
    for(int i = 0; i < n; height[Rank[i++]] = k){
        for(k?k--:0,j = sa[Rank[i]-1]; r[j+k] == r[i+k]; k++);
    }   
}
int ok(int m,int n){
    int num = 0;
    for(int i = 1; i <= n; i++){
        if(height[i] >= m){
            num ++;   
            if(num + 1 >= k)return 1;
        } else{
            num = 0;   
        }
    }
    return 0;
}
void solve(){
    int l,r,m,ans = 0;
    l = 0,r = n;
    while(l <= r){
        m = (l + r) >> 1;
        if(ok(m,n)){
            ans = max(m,ans);
            l = m + 1;
        } else {
            r = m - 1;   
        }
    }
    printf("%d\n",ans);
}
int main(){
    while(~scanf("%d%d",&n,&k)){
        for(int i = 0; i < n; i++){
            scanf("%d",&r[i]);
        }
        r[n] = 0;
        da(r,sa,n+1,1000002);
        calheight(r,sa,n);
        solve();
    }
    return 0;
}