\(\color{#0066ff}{ 题目描述 }\) 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…

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…

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…

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>…

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]\…

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,…

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…