poj3261 Milk Patterns 后缀数组+二分

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 the ith 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：给你一个串让你求k次可重复的最长子串的长度

思路：后缀数组的基础应用之一，首先长度应该是1~串长之间的数，所以自然而然想到二分结果，但二分结果后如何判断是否满足呢需要找到一个判断条件，这里就要用到height[]数组了，height数组表示相邻的后缀的最大前缀长度，所以在height中只要有连续k个最小值>=x,此时x就满足了条件

ac代码：

#include<iostream>
using namespace std;
const int MAX = 20050;
const int M = 20000;


int n, k, num[MAX];
int sa[MAX], rank[MAX], height[MAX];
int wa[MAX], wb[MAX], wv[MAX], wd[MAX];


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


void da(int *r, int n, int m)
{
    int i, j, p, *x = wa, *y = wb, *t;
    for(i = 0; i < m; i ++) wd[i] = 0;
    for(i = 0; i < n; i ++) wd[x[i]=r[i]] ++;
    for(i = 1; i < m; i ++) wd[i] += wd[i-1];
    for(i = n-1; i >= 0; i --) sa[-- wd[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 ++) wd[i] = 0;
        for(i = 0; i < n; i ++) wd[wv[i]] ++;
        for(i = 1; i < m; i ++) wd[i] += wd[i-1];
        for(i = n-1; i >= 0; i --) sa[-- wd[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 calHeight(int *r, 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 valid(int len)
{
    int i = 1, cnt;
    while(1)
    {
        while(i <= n && height[i] < len) i ++;
        if(i > n) break;
        cnt = 1;
        while(i <= n && height[i] >= len)
        {
            cnt ++;
            i++;
        }
        if(cnt >= k) return true;
    }
    return false;
}


int main()
{
    cin >> n >> k;
    for(int i = 0; i < n; i ++)
    {
        cin >> num[i];
    }
    num[n] = 0;
    da(num, n + 1, M);
    calHeight(num, n);


    int low = 1, high = n, mid;
    while(low <=high)
    {
        mid = (low+high) / 2;
        if(valid(mid))
        {
            low = mid+1;
        }
        else
        {
            high = mid - 1;
        }
    }
    cout << low-1<< endl;
    return 0;
}