SPOJ_705_New Distinct Substrings_后缀数组 题意: 给定一个字符串,求该字符串含有的本质不同的子串数量. 后缀数组的一个小应用. 考虑每个后缀的贡献,如果不要求本质不同那就是n-sa[i]-1. 然后需要去重,就是把height[i]这部分减掉. 代码: #include <stdio.h> #include <string.h> #include <algorithm> #include <stdlib.h> using n…

题目链接:https://vjudge.net/problem/SPOJ-SUBST1 SUBST1 - New Distinct Substrings #suffix-array-8 Given a string, we need to find the total number of its distinct substrings. Input T- number of test cases. T<=20; Each test case consists of one string, who…

Given a string, we need to find the total number of its distinct substrings. Input T- number of test cases. T<=20;Each test case consists of one string, whose length is <= 1000 Output For each test case output one number saying the number of distinc…

题意:统计母串中包含多少不同的子串 然后这是09年论文<后缀数组——处理字符串的有力工具>中有介绍 公式如下: 原理就是加上新的,减去重的,这题是因为打多校才补的,只能说我是个垃圾 #include <iostream> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <string> #include &l…

题意 : 对于给出的串,输出其不同长度的子串的种类数 分析 : 有一个事实就是每一个子串必定是某一个后缀的前缀,换句话说就是每一个后缀的的每一个前缀都代表着一个子串,那么如何在这么多子串or后缀的前缀中找出不同的并计数呢?思路就是所有的可能子串数 - 重复的子串数.首先我们容易得到一个长度为 len 的串的子串数为 len * ( len + 1) / 2.那如何知道重复的子串数呢?答案就是利用后缀数组去跑一遍 Height ,得到所有的最长公共前缀(LCP),这些最长公共前缀的值都存在了 He…

给定一个字符串,求不相同的子串的个数. 假如给字符串“ABA";排列的子串可能: A B A AB  BA ABA 共3*(3+1)/2=6种; 后缀数组表示时: A ABA BA 对于A和AB height[i]=1; 表明一个长度公共,所以ABA中多出现了A这个子串,所以6-1=5: 对于ABA BA height[i]=0,所以不需要减去. 最后答案为5: #include<iostream> #include<stdio.h> #include<string…

题意:求字符串中不同子串的个数. 解题关键:每个子串一定是某个后缀的前缀,那么原问题等价于求所有后缀之间的不相同的前缀的个数. 1.总数减去height数组的和即可. 注意这里height中为什么不需要进行组合计数,因为,每一个height的左端点已经确定,所以只需变动右端点,总共$height[i]$种情况. 2.如果所有的后缀按照 suffix(sa[1]), suffix(sa[2]),suffix(sa[3]), …… ,suffix(sa[n])的顺序计算,不难发现,对于每一次新加进来…

New Distinct Substrings 题意 给出T个字符串,问每个字符串有多少个不同的子串. 思路 字符串所有子串,可以看做由所有后缀的前缀组成. 按照后缀排序,遍历后缀,每次新增的前缀就是除了 与上一个后缀的所有公共前缀 之外的前缀. 答案就是用总数-重复的 即\(\frac{n(n+1)}{2}-\sum_{i=1}^{n}height[i]\) 代码 // #include <bits/stdc++.h> #include <stdio.h> #include &l…

题目链接:http://www.spoj.com/problems/DISUBSTR/ 思路: 每个子串一定是某个后缀的前缀,那么原问题等价于求所有后缀之间的不相同的前缀的个数.如果所有的后缀按照suffix(sa[1]),suffix(sa[2]),suffix(sa[3]),……suffix(sa[n])的顺序计算,不难发现,对于每一次新加进来的后缀suffix(sa[k]),它将产生n-sa[k]+1个新的前缀.但是其中有height[k]个是和前面的字符串的前缀是相同的.所以suffix…

Description Given a string, we need to find the total number of its distinct substrings. Input T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000 Output For each test case output one number saying the num…

Given a string, we need to find the total number of its distinct substrings. Input T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000 Output For each test case output one number saying the number of disti…

[SPOJ]Distinct Substrings/New Distinct Substrings(后缀数组) 题面 Vjudge1 Vjudge2 题解 要求的是串的不同的子串个数 两道一模一样的题目 其实很容易: 总方案-不合法方案数 对于串进行后缀排序后 不合法方案数=相邻两个串的不合法方案数的和 也就是\(height\)的和 所以\[ans=\frac{n(n+1)}{2}-\sum_{i=1}^{len}height[i]\] #include<iostream> #include…

Given a string, we need to find the total number of its distinct substrings. Input \(T-\) number of test cases. \(T<=20\); Each test case consists of one string, whose length is \(<=1000\) Output For each test case output one number saying the numbe…

Given a string, we need to find the total number of its distinct substrings. Input T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000 Output For each test case output one number saying the number of disti…

DISUBSTR - Distinct Substrings no tags  Given a string, we need to find the total number of its distinct substrings. Input T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 1000 Output For each test case outpu…

New Distinct Substrings(后缀数组) 给定一个字符串,求不相同的子串的个数.\(n<=50005\). 显然,任何一个子串一定是后缀上的前缀.先(按套路)把后缀排好序,对于当前的后缀\(S_i\),显然他有\(n-sa[i]\)个前缀.其中,有\(height[i]\)个前缀字符串在编号比它小的后缀中出现过,因此它对答案的贡献是\(n-sa[i]-height[i]\). #include <cstdio> #include <cstring> usin…

Distinct Substrings 题意 求一个字符串有多少个不同的子串. 分析 又一次体现了后缀数组的强大. 因为对于任意子串,一定是这个字符串的某个后缀的前缀. 我们直接去遍历排好序后的后缀字符串(也就是 \(sa\) 数组),每遍历到一个后缀字符串,会新添数量为这个后缀字符串的长度的前缀,但是要减去 \(height[i]\),即公共前缀的长度,因为前面已经添加过了这个数量的前缀串. code #include<bits/stdc++.h> using namespace std;…

Given a string, we need to find the total number of its distinct substrings. Input T- number of test cases. T<=20;Each test case consists of one string, whose length is <= 1000 Output For each test case output one number saying the number of distinc…

Spoj-DISUBSTR - Distinct Substrings New Distinct Substrings SPOJ - SUBST1 我是根据kuangbin的后缀数组专题来的 这两题题意一样求解字符串中不同字串的个数: 这个属于后缀数组最基本的应用 给定一个字符串,求不相同的子串的个数. 算法分析: 每个子串一定是某个后缀的前缀,那么原问题等价于求所有后缀之间的不相同的前缀的个数. 如果所有的后缀按照 suffix(sa[1]), suffix(sa[2]), suffix(sa…

[题目链接] http://www.spoj.com/problems/SUBST1/ [题目大意] 给出一个串,求出不相同的子串的个数. [题解] 对原串做一遍后缀数组,按照后缀的名次进行遍历, 每个后缀对答案的贡献为n-sa[i]+1-h[i], 因为排名相邻的后缀一定是公共前缀最长的, 那么就可以有效地通过LCP去除重复计算的子串. [代码] #include <cstdio> #include <cstring> #include <algorithm> usi…

给一个字符串求有多少个不相同子串. 每一个子串一定都是某一个后缀的前缀.由此可以推断出总共有(1+n)*n/2个子串,那么下面的任务就是找这些子串中重复的子串. 在后缀数组中后缀都是排完序的,从sa[1]到sa[n],这么思考以某个串为前缀的子串有几个,那么容易想到重复子串的个数其实就是∑height[i]. 所以结果就是(1+n)*n/2-∑height[i]. #include<cstdio> #include<cstring> #include<algorithm>…

题意: 问给定串有多少本质不同的子串? 思路: 子串必是某一后缀的前缀,假如是某一后缀\(sa[k]\),那么会有\(n - sa[k] + 1\)个前缀,但是其中有\(height[k]\)个和上一个重复,那么最终的贡献的新串为\(n - sa[k] + 1 - height[k]\).故最终结果为\(\sum_{i = 1}^n (n - sa[k] + 1 - height[k])\),即 \(\frac{n * (n + 1)}{2} - \sum_{i = 1}^nheight[k]\…

Description ?? is practicing his program skill, and now he is given a string, he has to calculate the total number of its distinct substrings. But ?? thinks that is too easy, he wants to make this problem more interesting. ?? likes a character X very…

Substring Problem Description ?? is practicing his program skill, and now he is given a string, he has to calculate the total number of its distinct substrings. But ?? thinks that is too easy, he wants to make this problem more interesting. ?? likes…

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…

2016暑假多校联合---Substring Problem Description ?? is practicing his program skill, and now he is given a string, he has to calculate the total number of its distinct substrings. But ?? thinks that is too easy, he wants to make this problem more interesti…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5008 Description In this problem, you are given a string s and q queries. For each query, you should answer that when all distinct substrings of string s were sorted lexicographically, which one is…

题目:http://cojs.tk/cogs/problem/problem.php?pid=1709 1709. [SPOJ705]不同的子串 ★★   输入文件:subst1.in   输出文件:subst1.out   简单对比时间限制:1 s   内存限制:256 MB [题目描述] 给定一个字符串,计算其不同的子串个数. [输入格式] 一行一个仅包含大写字母的字符串,长度<=50000 [输出格式] 一行一个正整数,即不同的子串个数. [样例输入] ABABA [样例输出] 9 [来源…

The Number of Palindromes Problem Description Now, you are given a string S. We want to know how many distinct substring of S which is palindrome. Input The first line of the input contains a single integer T(T<=20), which indicates number of test ca…

The Number of Palindromes Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2465    Accepted Submission(s): 841 Problem Description Now, you are given a string S. We want to know how many distin…