回文树模板（区间查询回文串数量）

题目描述

You've got a string s=s_{1}s_{2}...\ s_{|s|}s=s1​s2​... s∣s∣​ of length |s|∣s∣ , consisting of lowercase English letters. There also are qq queries, each query is described by two integers l_{i},r_{i}li​,ri​ (1<=l_{i}<=r_{i}<=|s|)(1<=li​<=ri​<=∣s∣) . The answer to the query is the number of substrings of string s[l_{i}...\ r_{i}]s[li​... ri​] , which are palindromes.

String s[l...\ r]=s_{l}s_{l+1}...\ s_{r}s[l... r]=sl​sl+1​... sr​ (1<=l<=r<=|s|)(1<=l<=r<=∣s∣) is a substring of string s=s_{1}s_{2}...\ s_{|s|}s=s1​s2​... 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}t=t1​t2​... t∣t∣​=t∣t∣​t∣t∣−1​... t1​ .

输入格式

The first line contains string ss (1<=|s|<=5000)(1<=∣s∣<=5000) . The second line contains a single integer qq (1<=q<=10^{6})(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}li​,ri​ (1<=l_{i}<=r_{i}<=|s|)(1<=li​<=ri​<=∣s∣) — the description of the ii -th query.

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

输出格式

Print qq 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 #1复制

caaaba
5
1 1
1 4
2 3
4 6
4 5

输出 #1复制

1
7
3
4
2

说明/提示

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

链接：https://codeforces.com/problemset/problem/245/H

题意：q次询问，【l,r】区间内多少回文串。

题解：dp打表回文树，q次询问用O(1)查询。

#include 
#include 
#include 
#include 
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1e6+5;
int dp[5005][5005];//非模板
struct PAM{
	
	int nex[maxn][30];//指向的串为当前串两端加上同一个字符构成
	int fail[maxn];//fail跳转到自己这个串的最长回文后缀 
	int cnt[maxn];//出现次数 
	int num[maxn];// 表示以节点i表示的最长回文串的最右端点为回文串结尾的回文串个数。
	int len[maxn];//len[i]表示节点i表示的回文串的长度 
	int S[maxn];//存放添加的字符
	int last,n,p;//last指向上一个字符所在的节点，方便下一次add
	
	int create(int rt){//新建节点 
		memset(nex[p],0,sizeof(nex[p]));
		cnt[p]=0;
		num[p]=0;
		len[p]=rt;
		return p++;
	}
	
	void init(){//初始化 
		p=last=n=0;
		create(0);
		create(-1);
		S[0]=-1;
		fail[0]=1;
	}
	
	int getFail(int x){//寻找失败节点 
		while(S[n-len[x]-1]!=S[n])	x=fail[x];
		return x;
	}
	void insert(int c)//插入字符 
	{
		c=c-'a';
		S[++n]=c;
		int cur=getFail(last);
		if(!nex[cur][c]){
			int now=create(len[cur]+2);
			fail[now]=nex[getFail(fail[cur])][c];
			nex[cur][c]=now;
			num[now]=num[fail[now]]+1;
		}
		last=nex[cur][c];
		cnt[last]++;
	}
	
	void count()//cnt答案不准确，需要调用更新一下。 
	{
		long long ans=0;
		for (int i = p-1; i >= 0; i--)
        	cnt[ fail[i] ] += cnt[i];
	}
	
	void set(string s,int x){//设置字符串 
		init();
		int len=s.size();
		for(int i=0;i>s;
	solve("#"+s);
	int q;
	scanf("%d",&q);
	while(q--)
	{
		int l,r;
		scanf("%d%d",&l,&r);
		cout<