POJ3261-Milk Patterns(后缀数组)

题意:给定一个整数序列，求这个整数序列中至少重复出现K次的子串的最大长度。

分析:我们求出lcp[]数组:相邻后缀的最长公共前缀，我们二分长度，然后判断是否存在一段连续的一组的长度>=二分的长度，并且至少存在k次。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
const int N = 20005;
int k;
int rk[N], tmp[N];
int sa[N], lcp[N];
int a[N];
int n, K;
bool compare_sa(int i, int j)
{
	if (rk[i] != rk[j]) return rk[i] < rk[j];
	else
	{
		int ri = i + k <= n ? rk[i + k] : -1;
		int rj = j + k <= n ? rk[j + k] : -1;
		return ri < rj;
	}	
}

void construct_sa(int a[], int sa[])
{
	for (int i = 0; i <= n; ++i)
	{
		sa[i] = i;
		rk[i] = i < n ? a[i] : -1;
	}
	for (k = 1; k <= n; k *= 2)
	{
		sort(sa, sa + n + 1, compare_sa);
		tmp[sa[0]] = 0;
		for (int i = 1; i <= n; ++i)
			tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
		for (int i = 0; i <= n; ++i)
			rk[i] = tmp[i];
	}
}

void construct_lcp(int a[], int sa[], int lcp[])
{
	for (int i = 0; i <= n; ++i) rk[sa[i]] = i;

	int h = 0;
	lcp[0] = 0;

	for (int i = 0; i < n; ++i)
	{
		int j = sa[rk[i] - 1];

		if (h > 0) --h;
		for (; j + h < n && i + h < n; ++h)
		{
			if (a[j + h] != a[i + h]) break;
		}
		lcp[rk[i] - 1] = h;
	}
}

bool check(int mid)
{
	int tot = 1;
	for (int i = 0; i < n; ++i)
	{
		if (lcp[i] >= mid)
			++tot;
		else
		{
			if (tot >= K)
				return true;
			else
				tot = 1;
		}
	}
	return tot >= K;
}

int main()
{	
	scanf("%d%d", &n, &K);

	for (int i = 0; i < n; ++i) scanf("%d", &a[i]);

	construct_sa(a, sa);
	construct_lcp(a, sa, lcp);

	int l = 0, r = n;

	while (l < r)
	{
		int mid = l + r + 1 >> 1;
		if (check(mid)) l = mid;
		else r = mid - 1;
	}

	printf("%d
", l);


	return 0;
}