POJ3261 Milks patterns(后缀数组)】的更多相关文章

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 dai…
题目链接:http://poj.org/problem?id=3261 思路: 后缀数组的很好的一道入门题目 先利用模板求出sa数组和height数组 然后二分答案(即对于可能出现的重复长度进行二分) ,二分的时候,对 height进行分组,看是否存在一组height值使得其重复的次数大于等于k 代码如下: #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> usin…
题目链接:https://vjudge.net/problem/POJ-3261 Milk Patterns Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 17157   Accepted: 7592 Case Time Limit: 2000MS Description Farmer John has noticed that the quality of milk given by his cows varies f…
                                                    Milk Patterns 题意:求可重叠的至少重复出现k次的最长的字串长. 这题的做法和上一题差不多,也是先二分答案,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.这个做法的时间复杂度为 O(nlogn). #include <map> #include <set> #include <q…
[USACO06FEC]Milk Patterns 题目描述: 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…
题目地址:http://poj.org/problem?id=3261 题目: 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…
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…
题目大意:求可重叠的相同子串数量至少是K的子串最长长度 洛谷传送门 依然是后缀数组+二分,先用后缀数组处理出height 每次二分出一个长度x,然后去验证,在排序的后缀串集合里,有没有连续数量多于K个串的长度>=x, 但据说有一种高端做法是把二分换成单调队列,能减少常数,可惜我并没有看懂...... 原题好像是哈希的骚操作,但网上的题解好像都是后缀数组...... 比上一道男人八题简单多了,我原来的错代码竟然卡过去了70分.. #include <cstdio> #include <…
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…
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…