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]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…