Milk Patterns 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 th…

题目链接 题目描述 给定一个字符串,求至少出现 \(k\) 次的最长重复子串,这 \(k\) 个子串可以重叠. 思路 二分 子串长度,据其将 \(h\) 数组 分组,判断是否存在一组其大小 \(\geq k\). Code #include <cstdio> #include <vector> #include <iostream> #define maxn 1000000 #define maxm maxn + 10 #define maxd 20010 using…

Milk Patterns   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…

题意 : 给出一个长度为 N 的序列,再给出一个 K 要求求出出现了至少 K 次的最长可重叠子串的长度 分析 : 后缀数组套路题,思路是二分长度再对于每一个长度进行判断,判断过程就是对于 Height 数组进行限定长度的分组策略,如果有哪一组的个数 ≥  k 则说明可行! 分组要考虑到一个事实,对于每一个后缀,与其相匹配能够产生最长的LCP长度的串肯定是在后缀数组中排名与其相邻. 一开始对分组的理解有误,所以想了一个错误做法 ==> 遍历一下 Height 将值 ≥ (当前二分长度) 的做一次贡…

题意 找出出现k次的可重叠的最长子串的长度 题解 用后缀数组. 然后求出heigth数组. 跑单调队列就行了.找出每k个数中最小的数的最大值.就是个滑动窗口啊 (不知道为什么有人写二分,其实写啥都差不多快,可能是因为二分是一个常见的模型吧) #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<algorithm> using namespac…

Milk Patterns Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7586   Accepted: 3448 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,…

Milk Patterns Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7938   Accepted: 3598 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,…

题目链接 题意:可重叠的 k 次最长重复子串.给定一个字符串,求至少出现 k 次的最长重复子串,这 k 个子串可以重叠. 分析:与POJ 1743做法类似,先二分答案,height数组分段后统计 LCP>=m 的子串的个数. #include <cstdio> #include <cstring> #include <algorithm> const int N = 2e4 + 5; int sa[N], rank[N], height[N]; int t[N],…

题意:给定一个字符串,求至少出现k 次的最长重复子串,这k 个子串可以重叠. 分析:经典的后缀数组求解题:先二分答案,然后将后缀分成若干组.这里要判断的是有没有一个组的符合要求的后缀个数(height[i] >= mid)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. #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using…

摘要:本文讨论了最长公共子串的的相关算法的时间复杂度,然后在后缀数组的基础上提出了一个时间复杂度为o(n^2*logn),空间复杂度为o(n)的算法.该算法虽然不及动态规划和后缀树算法的复杂度低,但其重要的优势在于可以编码简单,代码易于理解,适合快速实现. 首先,来说明一下,LCS通常指的是公共最长子序列(Longest Common Subsequence,名称来源参见<算法导论>原书第3版p223),而不是公共最长子串(也称为最长公共子串). 最长公共子串问题是在文本串.模式串中寻找共有的…