[SPOJ]Distinct Substrings(后缀自动机) 题面 Vjudge 题意:求一个串的不同子串的数量 题解 对于这个串构建后缀自动机之后 我们知道每个串出现的次数就是\(right/endpos\)集合的大小 但是实际上我们没有任何必要减去不合法的数量 我们只需要累加每个节点表示的合法子串的数量即可 这个值等于\(longest-shortest+1=longest-parent.longest\) #include<iostream> #include<cstdio&g…

[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…

[SPOJ]Distinct Substrings 求不同子串数量 统计每个点有效的字符串数量(第一次出现的) \(\sum\limits_{now=1}^{nod}now.longest-parents.longest\) My complete code #include<bits/stdc++.h> using namespace std; typedef long long LL; const LL maxn=3000; LL nod,last,n,T; LL len[maxn],fa…

694. Distinct Substrings Problem code: DISUBSTR   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 c…

Distinct Substrings Time Limit: 1000ms Memory Limit: 262144KB This problem will be judged on SPOJ. Original ID: DISUBSTR64-bit integer IO format: %lld      Java class name: Main   Given a string, we need to find the total number of its distinct subst…

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…

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…

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…

题目链接 题意:给定一个字符串,求不相同的子串的个数 分析:我们能知道后缀之间相同的前缀的长度,如果所有的后缀按照 suffix(sa[0]), suffix(sa[1]), suffix(sa[2]), …… ,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(sa[k])将“贡献” 出 n-sa[k]+1- heig…

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…

vjudge原地爆炸... 题意:求一个字符串不同的子串的个数 策略:后缀数组 利用后缀数组的sa和height两个功能强大的数组,我们可以实现上述操作 首先有个很显然的结论:一个字符串的所有子串=它后缀的所有前缀 这是很显然的,因为一个后缀的前缀遍历了所有以该后缀起点为起点的字符串的子串,那么如果我们遍历所有后缀的,就能找出这个字符串的所有子串了 所以对于一个起点为sa[i]的字符串,最多能提供的贡献就是l-sa[i]+1,而再考虑重复字符串的个数,也就是这个后缀所有的与其他后缀最长的公共前缀…

https://vjudge.net/problem/SPOJ-DISUBSTR 题意: 给定一个字符串,求不相同的子串的个数. 思路: #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<vector> #include<stack> #include<queue> #include<cmath>…

传送门 双倍经验(弱化版本) 考虑求出来heightheightheight数组之后用增量法. 也就是考虑每增加一个heightheightheight对答案产生的贡献. 算出来是∑∣S∣−heighti+1−sai\sum|S|-height_i+1-sa_i∑∣S∣−heighti​+1−sai​ 代码: #include<bits/stdc++.h> #define ri register int using namespace std; const int N=5e4+5; int n…

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…

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…

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…

705. New Distinct Substrings Problem code: SUBST1 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…

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…

DISUBSTR - Distinct Substrings 链接 题意: 询问有多少不同的子串. 思路: 后缀数组或者SAM. 首先求出后缀数组,然后从对于一个后缀,它有n-sa[i]-1个前缀,其中有height[rnk[i]]个被rnk[i]-1的后缀算了.所以再减去height[rnk[i]]即可. 代码: 换了板子. #include<cstdio> #include<algorithm> #include<cstring> #include<iostr…

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…

D - New Distinct Substrings 题目大意:求一个字符串中不同子串的个数. 裸的后缀数组 #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define pii pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using names…

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 <= 50000 Output For each test case output one number saying the number of disti…

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…

DISUBSTR - Distinct Substrings 题意:给你一个长度最多1000的字符串,求不相同的字串的个数. 思路:一个长度为n的字符串最多有(n+1)*n/2个,而height数组已经将所有的重复的都计算出来了,直接减去就行.需要注意的是在字符串的最后面加个0,不参与Rank排名,这样得到的height数组直接从1到n. char s[N]; int sa[N],Rank[N],height[N],c[N],t[N],t1[N],n,m; void build(int n) {…

SUBST1 - New 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 <= 50000 Output For each test case ou…

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…

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 distin…

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…