CF1073G.Yet Another LCP Problem
普及/提高-
通过率:0%
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题目描述
Let LCP(s,t) be the length of the longest common prefix of strings s and t . Also let s[x…y] be the substring of s from index x to index y (inclusive). For example, if $s = $ "abcde", then s[1…3]= "abc", s[2…5]= "bcde".
You are given a string s of length n and q queries. Each query is a pair of integer sets a1,a2,…,ak and b1,b2,…,bl . Calculate i=1∑i=kj=1∑j=lLCP(s[ai…n],s[bj…n]) for each query.
输入格式
The first line contains two integers n and q ( 1≤n,q≤2⋅105 ) — the length of string s and the number of queries, respectively.
The second line contains a string s consisting of lowercase Latin letters ( ∣s∣=n ).
Next 3q lines contains descriptions of queries — three lines per query. The first line of each query contains two integers ki and li ( 1≤ki,li≤n ) — sizes of sets a and b respectively.
The second line of each query contains ki integers a1,a2,…aki ( 1≤a1<a2<⋯<aki≤n ) — set a .
The third line of each query contains li integers b1,b2,…bli ( 1≤b1<b2<⋯<bli≤n ) — set b .
It is guaranteed that i=1∑i=qki≤2⋅105 and i=1∑i=qli≤2⋅105 .
输出格式
Print q integers — answers for the queries in the same order queries are given in the input.
输入输出样例
输入#1
7 4 abacaba 2 2 1 2 1 2 3 1 1 2 3 7 1 7 1 1 2 3 4 5 6 7 2 2 1 5 1 5
输出#1
13 2 12 16
说明/提示
Description of queries:
- In the first query s[1…7]=abacaba and s[2…7]=bacaba are considered. The answer for the query is LCP(abacaba,abacaba)+LCP(abacaba,bacaba)+LCP(bacaba,abacaba)+LCP(bacaba,bacaba)=7+0+0+6=13 .
- In the second query s[1…7]=abacaba , s[2…7]=bacaba , s[3…7]=acaba and s[7…7]=a are considered. The answer for the query is LCP(abacaba,a)+LCP(bacaba,a)+LCP(acaba,a)=1+0+1=2 .
- In the third query s[1…7]=abacaba are compared with all suffixes. The answer is the sum of non-zero values: LCP(abacaba,abacaba)+LCP(abacaba,acaba)+LCP(abacaba,aba)+LCP(abacaba,a)=7+1+3+1=12 .
- In the fourth query s[1…7]=abacaba and s[5…7]=aba are considered. The answer for the query is LCP(abacaba,abacaba)+LCP(abacaba,aba)+LCP(aba,abacaba)+LCP(aba,aba)=7+3+3+3=16 .