【CF245H】Queries for Number of Palindromes(字符串区间dp)
Queries for Number of Palindromes - 洛谷
# Queries for Number of Palindromes
## 题面翻译
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 $ q $ 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 $ t $ 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} $ .
## 输入格式
The first line contains string $ s $ $ (1<=|s|<=5000) $ . The second line contains a single integer $ q $ $ (1<=q<=10^{6}) $ — the number of queries. Next $ q $ lines contain the queries. The $ i $ -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.
## 样例 #1
### 样例输入 #1
```
caaaba
5
1 1
1 4
2 3
4 6
4 5
```
### 样例输出 #1
```
1
7
3
4
2
```
## 提示
Consider the fourth query in the first test case. String $ s[4...\ 6] $ = «aba». Its palindrome substrings are: «a», «b», «a», «aba».
核心思路
类似容斥dp
f[l][r]表示答案 = f[l][r-1] + [l+1][r] -f[l+1][r-1]+pd[l][r]
AC代码
#include
using namespace std;
const int maxn = 5050;
int tot;
string a;
int n;
bool pd[5050][5050];
long long f[5050][5050];
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cin>>a;
n = a.size();
a = " "+a;
for(int i = 1;i <= n;i++)pd[i][i] = 1;
for(int i = 1;i <= n-1;i++)if(a[i] == a[i+1])pd[i][i+1] = 1;
for(int i = 3;i <= n;i++){
for(int l = 1,r = l+i-1;r <= n;l++,r++){
pd[l][r] = (pd[l+1][r-1] == 1&&a[l] == a[r]?1:0);
}
}
for(int i = 1;i <= n;i++){
f[i][i] = 1;
}
for(int i = 2;i <= n;i++){
for(int l = 1,r = l+i-1;r <= n;l++,r++){
f[l][r] = f[l][r-1]+f[l+1][r]-f[l+1][r-1]+pd[l][r];
}
}
int q;
cin>>q;
while(q--){
int l,r;
cin>>l>>r;
cout<