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…

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

\(\color{#0066ff}{ 题目描述 }\) 给定一个字符串,求该字符串含有的本质不同的子串数量. \(\color{#0066ff}{输入格式}\) T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000 \(\color{#0066ff}{输出格式}\) For each test case output one number saying th…

SAM里的转台不会有重复串,所以答案就是每个right集合所代表的串个数的和 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=100005; int T,n,fa[N],ch[N][27],dis[N],cur=1,con=1,la; long long ans; char s[N]; void ins(int c,int id) { la=cu…

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]\…

将所有后缀按照字典序排序后,每新加进来一个后缀,它将产生n - sa[i]个前缀.这里和小罗论文里边有点不太一样. height[i]为和字典序前一个的LCP,所以还要减去,最终累计n - sa[i] - height[i]即可. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; char s[maxn]; int sa[maxn], rank[maxn]…

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>…

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…

SUBST1 - New Distinct Substrings 和上一题题意一样,只是数据范围有所改动,50000. 思路还是和上一题一样,所有字串数(len+1)*len/2.注意这里可能爆int,所有需要处理一下,然后减去height数组. char s[N]; int sa[N],Rank[N],height[N],c[N],t[N],t1[N],n,m; void build(int n) { // printf("n=%d m=%d\n",n,m); int i,*x=t,…

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…

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 <= 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…

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…

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…

[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…

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…

[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…

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…

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;…

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) {…

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…

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…

[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 [思路] 一看就知道是后缀数组题啦-但是我不会写QAQ..只好现学现用啦- 在字符串最后补上一个'$',不因为别的只因为它比‘A’还要小..不然你补ascII码是0的也可以.. 申请rank数组和sa数组,rank[i]=j代表后缀i排第j位,sa[i]=j代表排名第i的是后缀j.也就…

写在前面:因为要准备面试,开始了在[LeetCode]上刷题的历程.LeetCode上一共有大约150道题目,本文记录我在<http://oj.leetcode.com>上AC的所有题目,以Leetcode上AC率由高到低排序,基本上就是题目由易到难.我应该会每AC15题就过来发一篇文章,争取早日刷完.所以这个第二篇还是相对比较简单的15道题了. 部分答案有参考网上别人的代码,和leetcode论坛里的讨论,很多答案肯定有不完美的地方,欢迎提问,指正和讨论. No.16 Climbing St…