CF245H Queries for Number of Palindromes（回文树）

题意翻译

题目描述

给你一个字符串s由小写字母组成，有q组询问，每组询问给你两个数，l和r，问在字符串区间l到r的字串中，包含多少回文串。

输入格式

第1行，给出s，s的长度小于5000 第2行给出q(1<=q<=10^6) 第2至2+q行 给出每组询问的l和r

输出格式

输出每组询问所问的数量。

题目描述

You've got a string s=\(s_{1}\)\(s_{2}\)...\(s_{|s|}\) of length |s| , consisting of lowercase English letters. There also are qq queries, each query is described by two integers \(l_{i}\),\(r_{i}\) (1<=\(l_{i}\)<=\(r_{i}\)<=|s|) . The answer to the query is the number of substrings of string \(s[\) \(l_{i}\) ... \(r_{i}\) \(]\) , which are palindromes.

String \(s[l...\ r]\)=\(s_{l}\)\(s_{l+1}\)...\(\ s_{r}\)(1<=l<=r<=∣s∣) is a substring of string \(s=s_{1}s_{2}...\ s_{|s|}\) .

String tt is called a palindrome, if it reads the same from left to right and from right to left. Formally, if \(t=t_{1}t_{2}...\ t_{|t|}=t_{|t|}t_{|t|-1}...\ t_{1}\).

输入输出格式

输入格式：

The first line contains string ss (1<=|s|<=5000) . The second line contains a single integer qq (1<=q<=10⁶) — the number of queries. Next qq lines contain the queries. The ii -th of these lines contains two space-separated integers \(l_{i}\),\(r_{i}\) (1<=\(l_{i}\)<=\(r_{i}\)<=|s|) — the description of the i-th query.

It is guaranteed that the given string consists only of lowercase English letters.

输出格式：

Print q integers — the answers to the queries. Print the answers in the order, in which the queries are given in the input. Separate the printed numbers by whitespaces.

输入输出样例

输入样例#1： 复制

caaaba

5

1 1

1 4

2 3

4 6

4 5

输出样例#1： 复制

1

7

3

4

2

说明

Consider the fourth query in the first test case. String \(s[4...\ 6]\) = «aba». Its palindrome substrings are: «a», «b», «a», «aba».

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题解

这个题目的思路非常巧妙？

因为时间复杂度允许达到\(n\)²,于是我们就从1开始一直到strlen(s)，\(l\)的位置每向后移动一位就清空回文树，并把这个\(l...len\)的回文串重新放入回文树。用\(ans[l][r]\)来统计一下答案

然后o(1)查询就OK了。

然后我yy了一下莫队？

是不是离线的话时间复杂度就降到了\(n{\sqrt n}\)了呢？

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代码

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
    int fail,len,ch[26],dep;
}t[5001];
int tot,k,ans[5001][5001];
char s[5001],ch[5001];
int read()
{
    int x=0,w=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*w;
}

void clear()
{
    memset(t,0,sizeof(t));
    tot=1;k=0;t[0].fail=t[1].fail=1;t[1].len=-1;
}

void solve()
{
    int n=strlen(ch+1);
    for(int i=1;i<=n;i++)
    {
        clear();
        for(int j=i;j<=n;j++)s[j-i+1]=ch[j];
        for(int j=i;j<=n;j++)
        {
            //clear();
            while(s[j-i+1-t[k].len-1]!=s[j-i+1])k=t[k].fail;
            if(!t[k].ch[s[j-i+1]-'a']){
                t[++tot].len=t[k].len+2;
                int l=t[k].fail;
                while(s[j-i+1-t[l].len-1]!=s[j-i+1])l=t[l].fail;
                t[tot].fail=t[l].ch[s[j-i+1]-'a'];
                t[k].ch[s[j-i+1]-'a']=tot;
                t[tot].dep=t[t[tot].fail].dep+1;
        }
            k=t[k].ch[s[j-i+1]-'a'];
            ans[i][j]=ans[i][j-1]+t[k].dep;
        }
    }
}

int main()
{
    scanf("%s",ch+1);
    solve();
    int q=read();
    for(int i=1;i<=q;i++)
    {
    int l=read(),r=read();
    printf("%d\n",ans[l][r]);
    }
    return 0;
}