传送门 解题思路 感觉这种题都是套路之类的??首先把三个串并成一个,中间插入一些奇怪的字符,然后跑遍\(SA\).考虑按照\(height\)分组计算,就是每个\(height\)只在最高位计算一次,然后求个后缀和,这个可以并查集来实现.具体就是记一个\(sum[i][3]\)表示第\(i\)个联通块中\(0,1,2\)的个数,\(0,1,2\)就是出现在三个串的哪一个,然后合并时需要容斥一下. 代码 #include<iostream> #include<cstdio> #inc

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28859   Accepted: 12045 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "

连续重复子串问题 poj 2406 Power Strings http://poj.org/problem?id=2406 问一个串能否写成a^n次方这种形式. 虽然这题用kmp做比较合适,但是我们还是用后缀数组做一做,巩固后缀数组的能力. 对于一个串,如果能写出a^n这种形式,我们可以暴力枚举循环节长度L,那么后缀suffix(1)和suffix(1 + L)的LCP应该就是 lenstr - L.如果能满足,那就是,不能,就不是. 这题的话da算法还是超时,等我学了DC3再写上来. 其实这

[Codechef SSTORY] A Story with Strings Description 给定两个字符串,求它们的最长公共子串.如果解不唯一,输出最先在第二个字符串中出现的那个. Solution 仍然考虑二分答案,然后每次检验连续的一块中是否有来自两个串的后缀出现. 最后额外处理一次,每次检验连续的一块中是否有来自两个串的后缀出现,在有的情况下利用所有来源于第二个串的后缀得到可能的最小位置即可. 竟然二分边界忘记加一,我TM真是个憨憨 #include <bits/stdc++.h

Power Strings 题意 给出一个字符串s,求s最多由几个相同的字符串重复而成(最小循环节的重复次数) 思路 之前学习KMP的时候做过. 我的思路是:枚举字符串的长度,对于当前长度k,判断\(lcp(1,k+1)>=k\),\(lcp(k+1,2k+1)>=k\),\(lcp(3k+1,4k+1)>=k\)....是否都成立 但这样复杂度有点高,就找了一下题解. 其实只需要判断\(lcp(1,k+1)\)是否是\(n-k\)就可以了. 具体原理如下图: 参考博客 对于\(lcp(

http://poj.org/problem?id=2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27003   Accepted: 11311 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = &q

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defin

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defin

题目链接:https://vjudge.net/problem/POJ-2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 52631   Accepted: 21921 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc"

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defin

这题可以用后缀数组,KMP方法做 后缀数组做法开始想不出来,看的题解,方法是枚举串长len的约数k,看lcp(suffix(0), suffix(k))的长度是否为n- k ,若为真则len / k即为结果. 若lcp(suffix(0), suffix(k))的长度为n- k,则将串每k位分成一段,则第1段与第2段可匹配,又可推得第2段与第3段可匹配……一直递归下去,可知每k位都是相同的,画图可看出匹配过程类似于蛇形. 用倍增算法超时,用dc3算法2.5秒勉强过. #include<cstdi

后缀数组专题的 emm.. 就next 循环节../ 有后缀数组也可以做 从小到大枚举长度i,如果长度i的子串刚好是重复了len/i次,应该满足len % i == 0和rank[0] - rank[i] == 1(整个串的等级比 i位置开始的后缀的等级大1  (i位置开始的后缀即为比总串低一个等级的后缀)) 和height[rank[0]] == len-i (整个串 和 比它低一个等级的串的最长公共前缀的长度 是总长度减去这个循环节的长度)这些条件的 #include <iostream>

电源串 时间限制: 3000MS   内存限制: 65536K 提交总数: 53037   接受: 22108 描述 给定两个字符串a和b,我们定义a * b是它们的连接.例如,如果a ="abc"和b ="def",那么a * b ="abcdef".如果我们将连接看作是乘法,则用正常的方式定义非负整数的指数:a ^ 0 =""(空字符串)和a ^(n + 1)= a *(a ^ n). 输入 每个测试用例都是一行代表s的输

题意: n<=1000000,cas较大 思路:这是一道论文题 后缀数组已弃疗,强行需要DC3构造,懒得(不会)写 ..]of longint; n,m,i,j,len,ans,st:longint; ch:ansistring; flag:boolean; function min(x,y:longint):longint; begin if x<y then exit(x); exit(y); end; function cmp(a,b,l:longint):boolean; begin

题意:求最小循环节循环的次数. 题解:这个题其实可以直接用kmp去求最小循环节,然后在用总长度除以循环节.但是因为在练后缀数组,所以写的后缀数组版本.用倍增法会超时!!所以改用DC3法.对后缀数组还不是很理解,找了很多博客也没看懂到底有些数组到底记录的是啥,但他的实现过程很好理解,等我弄懂了再来给博客加注释吧. 先求出sa数组,height数组,rank数组(因为和c++库中某个东西重了所以写成rnk数组),数组一定要开3倍.接下来从小到大枚举循环节长度 i,如果长度i的子串刚好是重复了len/

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=5558 Problem Description Alice wants to send a classified message to Bob. She tries to encrypt the message with her original encryption method. The message is a string S, which consists of Nlowercase let

Long Long Message   Description The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a bi

后缀数组被称为字符串处理神器,要解决字符串问题,一定要掌握它.(我这里的下标全部都是从1开始) 首先后缀数组要处理出两个数组,一个是sa[],sa[i]表示排名第i为的后缀的起始位置是什么,rank[i]表示第i个字符为起始点的后缀,它的排名是什么.可以知道sa[rank[i]] = i; rank[sa[i]] = i; 由于每个后缀各不相同,至起码长度不同,所以每个后缀是不可能相等的. 解除一个值,就能在O(n)时间内得到另外一个. 定义:suffix(i)表示从[i, lenstr]这个后

Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 26601   Accepted: 10816 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

Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 24756   Accepted: 10130 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: his mother is ge