题目链接: Palindrome Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 113    Accepted Submission(s): 44 Problem Description Alice like strings, especially long strings. For each string, she has a s…

Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string?" A string is sai…

Palindrome Time Limit: 15000MS   Memory Limit: 65536K Total Submissions: 12616   Accepted: 4769 Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you pr…

题目大意就是说在给定的字符串里找出一个长度最大的回文子串. 才开始接触到manacher,不过这个算法的确很强大,这里转载了一篇有关manacher算法的讲解,可以去看看:地址 神器: #include <map> #include <set> #include <stack> #include <queue> #include <cmath> #include <ctime> #include <vector> #inc…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud 求最长回文子串. http://acm.timus.ru/problem.aspx?space=1&num=1297 Manacher模板题,复杂度O(n),做这题纯属是为了验一下自己写的模板是否正确. 当然这题也可以用后缀数组来搞 #include <iostream> #include <sstream> #include <ios> #in…

题链: http://poj.org/problem?id=3974 题解: Manacher 求最长回文串长度. 终于会了传说中的马拉车,激动.推荐一个很棒的博客:https://www.61mon.com/index.php/archives/181/ 代码: #include<cstdio> #include<cstring> #include<iostream> #define MAXN 2005000 #define filein(x) freopen(#x&…

不知道错在哪了,求大神指教!!! 思路:用manacher求出每个以str[i]为中心轴的回文串的长度,RMQ预处理区间最大值,对于每个查询,二分最大回文串长,判定是否可行. #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; ; char str[ MAXN ]; ]; ]; ][]; int n, len; vo…

这道题可以$O(nlogn)$,当然也可以$O(n)$做啦$qwq$ $O(nlogn)$的思路是枚举每个回文中心,通过哈希预处理出前缀和后缀哈希值备用,然后二分回文串的长度,具体的就是判断在长度范围内,前缀哈希值和后缀哈希值是否相等. #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<ccty…

题面 题意:给你一个字符串,问你满足s[i]=s[2n-i]=s[2n+i-2]的子串(这子串长度为3n-2)有多少个,原字符串长度<=5e5 题解:对于这种子串,其实要满足2个回文,跑过一次Manacher后,len[i]表示以i向两边扩展最远的回文串长度, 那么对于答案,实际就是统计满足下列条件(i,j)的对数 i  <= j j - i <= len[i] j - i <= len[j] 移项就是 i >= j -  len[j] j <= i + len[i]…

之前用字符串hash+二分过了,今天刚看了manacher拿来试一试. 这manacher也快太多了%%% #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; ; *maxn]; ], id, mx, ans; inline void manacher(){ mx = id = ans = ; memset( p,…

题意 : 给定一个字符串S,问你有多少长度为 n 的子串满足  S[i]=S[2n−i]=S[2n+i−2] (1≤i≤n) 参考自 ==> 博客 分析 : 可以看出满足题目要求的特殊回文子串其实是根据 n 以及 2*n-1 对称的,如图所示 如果我设第一个对称点为 i 第二个为 j ,p[i] 为以 i 为中心是回文半径, 那满足题目条件的子串必定满足  j − p[j] + 1 ≤ i < j ≤ i + p[i] − 1 简单点说就是两点的回文半径要相互覆盖 p[]数组很容易使用 Man…

E. Palindrome QueryTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100570/problem/E Description De Prezer loves palindrome strings. A string s1s2...sn is palindrome if and only if it is equal to its reverse. De Prezer also love…

http://poj.org/problem?id=3974 Palindrome Time Limit: 15000MSMemory Limit: 65536K Total Submissions: 2707Accepted: 995 Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a simpl…

Palindrome Time Limit: 15000MS   Memory Limit: 65536K Total Submissions: 6183   Accepted: 2270 Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you pro…

 Palindrome Time Limit:15000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, &quo…

题目链接: http://codeforces.com/problemset/problem/7/D D. Palindrome Degree time limit per test1 secondmemory limit per test256 megabytes 问题描述 String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length…

Palindrome Time Limit: 15000MS   Memory Limit: 65536K Total Submissions: 14021   Accepted: 5374 题目链接:http://poj.org/problem?id=3974 Description: Andy the smart computer science student was attending an algorithms class when the professor asked the st…

1297. Palindrome Time limit: 1.0 second Memory limit: 64 MB The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots»…

D. Palindrome Degree time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output String s of length n is called k-palindrome, if it is a palindrome itself, and its prefix and suffix of length  are(k - …

马拉车用于解决最长回文子串问题,重点是子串,而不是子序列,时间复杂度为O(n). 解释一下变量的意义: Len[i]数组去存第i个位置到mx位置的长度 id记录上一次操作的位置(这个操作可以看模板) mx标记上一次的最长子串的最右端 模板: 1 void init() //这个是用来处理字符串的 2 { 3 memset(str,0,sizeof(str)); 4 int k=0; 5 str[k++]='$'; 6 for(int i=0;i<len;++i) 7 str[k++]='#',s…

This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task. You are given a string ss, consisting of lowercas…

寻找字符串中的最长回文序列和所有回文序列(正向和反向一样的序列,如aba,abba等)算是挺早以前提出的算法问题了,最近再刷Leetcode算法题的时候遇到了一个(题目),所以就顺便写下. 如果用正反向遍历的方法的话时间复杂度将会是O(N^2),而利用Manacher算法将会是O(N),在处理长序列的时候能显著提高速度. 算法原理 回文序列的左右是对称的,也就是说在找到一个回文序列的时候,回文序列的右半部份将会是左半部分的镜像,在符合一定条件的时候可以直接判断以当前字符为中心的回文序列的长度 以…

多敲几个模板题,加深一下对Manacher算法的理解. 这道题给的时间限制15s,是我见过的最长的时间的了.看来是为了让一些比较朴素的求最大回文子串的算法也能A过去 Manacher算法毕竟给力,运行时间200+MS //#define LOCAL #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; ];…

1297. Palindrome Time limit: 1.0 secondMemory limit: 64 MB The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots» s…

题意:就是求一个串的最长回文子串....输出长度. 直接上代码吧,没什么好分析的了.   代码如下: ============================================================================================================================== #include<stdio.h> #include<string.h> #include<algorithm> us…

传送门 假设字符串\(B,D\)满足\(|B| \geq |D|\),那么一定会有\(B=rev(D)+T\),其中\(T\)是一个回文串. 考虑枚举回文串\(T\)的中心\(p\),找到以\(p\)为中心的最长回文串\(S[l,r]\).值得注意的是,回文串越长一定越好,因为如果回文串不是最长的,那么可以向左右拓展一位,\(B\)串长度会\(+1\),\(D\)串长度会\(-1\),没有变化.所以我们直接Manacher求最长回文串即可. 现在我们需要求的就是最小的\(i\)满足\(rev(S…

http://poj.org/problem?id=3974 模板题,Manacher算法主要利用了已匹配回文串的对称性,对前面已匹配的回文串进行利用,使时间复杂度从O(n^2)变为O(n). https://www.cnblogs.com/xiaoningmeng/p/5861154.html 详细解释 https://www.zhihu.com/question/30226229 这是复杂度O(n)的解释 代码 #include<cstdio> #include<cstring>…

The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots» security service would have already started an undercover op…

Manacher算法教程:http://blog.csdn.net/ggggiqnypgjg/article/details/6645824 模板题,Code 附带注释: #include<cstdio> #include<algorithm> #include<cstring> using namespace std; ],a[]; ]; ],zu;//f[i]表示插入一堆#后,以i为中心的最长回文子串の半径长度, //故其减1后就是原串的最长回文子串的答案 //也是…

https://nanti.jisuanke.com/t/15428 题目大意:离散表示的字符串,求其最长回文串长度. 解题关键:若只用Manacher算法,在统计sum时会超时,所以加一个树状数组来维护前n项和,即可AC. 注意进行Manacher时,i是从1开始的,不要小也不要大. n天后更新:以前一直没搞清离线与动态维护的区别,今天终于理解了,此题可以开始直接离线求前n项和,因为以后不再改变,所以此题不需要bit维护.直接用一个数组统计一下前n项和即可. #include<bits/std…