POJ-3261-Milk Patterns-二分+哈希

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POJ 3261 Milk Patterns （求可重叠的k次最长重复子串）+后缀数组模板

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,…

POJ 3261 Milk Patterns 【后缀数组最长可重叠子串】

题目题目:http://poj.org/problem?id=3261 Milk Patterns Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 18880 Accepted: 8336 Case Time Limit: 2000MS Description Farmer John has noticed that the quality of milk given by his cows varies from day t…

Poj 3261 Milk Patterns(后缀数组+二分答案)

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…

POJ 3261 Milk Patterns 可重复k次的最长重复子串

Milk PatternsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=3261 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 c…

POJ 3261 Milk Patterns 后缀数组求一个串种最长可重复子串重复至少k次

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…

poj 3261 Milk Patterns（后缀数组）(k次的最长重复子串)

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,…

POJ 3261 Milk Patterns （后缀数组，求可重叠的k次最长重复子串）

Milk Patterns Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 16742 Accepted: 7390 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,…

POJ 3261 Milk Patterns（后缀数组+二分答案）

[题目链接] http://poj.org/problem?id=3261 [题目大意] 求最长可允许重叠的出现次数不小于k的子串. [题解] 对原串做一遍后缀数组,二分子串长度x,将前缀相同长度超过x的后缀分组, 如果存在一个大小不小于k的分组,则说明答案可行,分治得到最大可行解就是答案. [代码] #include <cstdio> #include <cstring> #include <vector> using namespace std; const int…

poj 3261 Milk Patterns 后缀数组 + 二分

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…

POJ 3261 Milk Patterns（后缀数组+二分答案+离散化）

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…

后缀数组 POJ 3261 Milk Patterns

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],…

[poj 3261]Milk Patterns

后缀数组搞一下就可以了喵~ 其实这道题的第一个想法是 SAM ,建完后缀自动机后拓扑排序跑一遍统计下每个子串的出现次数就 O(N) 就妥妥过掉了后缀树也是 O(N) 的,统计一下每个节点对应的子树中有多少个叶子节点即可可是看了一下数据范围,欸呦你妹啊,字符数 1000000+1 是什么心态啊!居然把我的后缀树和后缀 zdj 都卡掉了啊! 然后很痛苦的想了想后缀数组的做法: 搞出 sa, rank, height 三个数组后,跑一遍二分,用一个 tot 表示找到了几个子串二分答案 L,O(N…

POJ 3261 Milk Patterns(后缀数组+单调队列)

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…

POJ 3261 Milk Patterns ( 后缀数组 && 出现k次最长可重叠子串长度 )

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…

POJ 2774 后缀数组 || 二分+哈希

Long Long Message Time Limit: 4000MS Memory Limit: 131072K Total Submissions: 35607 Accepted: 14275 Case Time Limit: 1000MS Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days…

POJ 3261 (后缀数组二分) Milk Patterns

这道题和UVa 12206一样,求至少重复出现k次的最长字串. 首先还是二分最长字串的长度len,然后以len为边界对height数组分段,如果有一段包含超过k个后缀则符合要求. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; + ; int s[maxn]; int sa[maxn], height[maxn], rank[maxn]; int t[m…

【POJ】3261 Milk Patterns

http://poj.org/problem?id=3261 题意:一个长度为n的串,要求最长的子串的长度且这个子串的出现次数不少于k次.(1<=n<=20000, 2<=k<=n) #include <cstdio> #include <algorithm> using namespace std; const int N=20015; void sort(int *x, int *y, int *sa, int n, int m) { static in…

poj 2002 Squares 几何二分 || 哈希

Squares Time Limit: 3500MS Memory Limit: 65536K Total Submissions: 15137 Accepted: 5749 Description A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating abou…

POJ 3261 后缀数组+二分

思路: 论文题- 二分+对后缀分组这块一开始不用基数排序会更快的(其实区别不大) //By SiriusRen #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define N 20050 int n,k,s[N],cntA[N],cntB[N],A[N],B[N],rk[N],sa[N],tsa[N],ht[N]; struct Node{int pos…

【poj 3261】Milk Patterns 后缀数组

Milk Patterns 题意给出n个数字,以及一个k,求至少出现k次的最长子序列的长度思路和poj 1743思路差不多,二分长度,把后缀分成若干组,每组任意后缀公共前缀都>=当前二分的长度.统计是否有某个组后缀数量>=k,如果有当前长度就可以. 代码 // #include <bits/stdc++.h> #include <stdio.h> #include <algorithm> #include <string.h> #defin…

BZOJ 1717: [Usaco2006 Dec]Milk Patterns 产奶的模式( 二分答案 + 后缀数组 )

二分答案m, 后缀数组求出height数组后分组来判断. ------------------------------------------------------------ #include<bits/stdc++.h> using namespace std; const int maxn = 20009; struct HASH { int id[maxn], N; HASH() { N = 0; } inline void work() { sort(id, id +…

[bzoj1717][Usaco2006 Dec]Milk Patterns 产奶的模式_后缀数组_二分答案

Milk Patterns 产奶的模式 bzoj-1717 Usaco-2006 Dec 题目大意:给定一个字符串,求最长的至少出现了$k$次的子串长度. 注释:$1\le n\le 2\cdot 10^4$,$2\le k\le n$. 想法:不难想到二分答案,现在我们考虑如何验证. 这里就是后缀数组的一个妙用了. 我们对原串建立后缀数组,观察$ht$数组. 考虑当前二分出来的$mid$.如果有至少连续$k$的$ht$值都不小于$mid$,那么$k$就是合法的. 故此我们直接扫$ht$数组看看…