2040. Palindromes and Super Abilities 2 题目连接: http://acm.timus.ru/problem.aspx?space=1&num=2040 Description Dima adds letters s1, -, sn one by one to the end of a word. After each letter, he asks Misha to tell him how many new palindrome substrings a…
Palindromes and Super Abilities 2 Time Limit: 1MS   Memory Limit: 102400KB   64bit IO Format: %I64d & %I64u Status Description Dima adds letters s1, -, sn one by one to the end of a word. After each letter, he asks Misha to tell him how many new pali…
Palindromes and Super Abilities 2 题目链接: http://acm.hust.edu.cn/vjudge/contest/126823#problem/E Description Dima adds letters s 1, -, s n one by one to the end of a word. After each letter, he asks Misha to tell him how many new palindrome substrings…
Palindromes and Super Abilities 2Time Limit: 500MS Memory Limit: 102400KB 64bit IO Format: %I64d & %I64u Description Dima adds letters s1, …, sn one by one to the end of a word. After each letter, he asks Misha to tell him how many new palindrome sub…
 Palindromes and Super Abilities Problem's Link: http://acm.timus.ru/problem.aspx?space=1&num=1960 Mean: 给你一个长度为n的字符串S,输出S的各个前缀中回文串的数量. analyse: 回文树(回文自动机)的模板题. 由于回文自动机中的p是一个计数器,也相当于一个指针,记录的是当前插入字符C后回文树中有多少个节点. 那么我们只需要一路插,一路输出p-2就行. p-2是因为一开始回文树中就有两个…
Palindromes and Super Abilities Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on Ural. Original ID: 196064-bit integer IO format: %lld      Java class name: (Any) After solving seven problems on Timus Online Judge with a word “…
http://acm.timus.ru/problem.aspx?space=1&num=1960 题意:给一个串s,要求输出所有的s[0]~s[i],i<|s|的回文串数目.(|s|<=10^5) #include <bits/stdc++.h> using namespace std; struct PT { static const int nS=26, nL=100015, N=nL; int f[N], l[N], c[N][nS], id[N], s[nL],…
Problem Palindromes and Super Abilities 2 (URAL2040) 题目大意 给一个字符串,从左到右依次添加,询问每添加一个字符,新增加的回文串数量. 解题分析 用回文自动机来做,如果新添加了一个字符,自动机中新开了一个节点,说明新增加了一个回文串. 对于每新添加一个字符,新增加的回文串数量最多为1. 另外,这道题既卡空间又卡时间,交了n发才过. 参考程序 #include <map> #include <set> #include <s…
After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Misha has got an unusual ability. Now, when he reads a word, he can mentally count the number of unique nonempty substrings of this word that are palindr…
写题遇上一棘手的题,[Apio2014]回文串,一眼看过后缀数组+Manacher.然后就码码码...过是过了,然后看一下[Status],怎么慢这么多,不服..然后就搜了一下,发现一种新东西——回文树. 回文树的功能: 假设我们有一个串S,S下标从0开始,则回文树能做到如下几点: 1.求串S前缀0~i内本质不同回文串的个数(两个串长度不同或者长度相同且至少有一个字符不同便是本质不同) 2.求串S内每一个本质不同回文串出现的次数 3.求串S内回文串的个数(其实就是1和2结合起来) 4.求以下标i…