CodeForces - 245H Queries for Number of Palindromes

区间型DP求解回文串数量

最新推荐文章于 2021-10-25 19:31:52 发布

转载最新推荐文章于 2021-10-25 19:31:52 发布 · 152 阅读

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原文链接：http://www.cnblogs.com/Loi-Vampire/p/6017047.html

本文介绍了一种使用区间动态规划方法解决给定字符串区间内回文子串数量的问题。通过预处理得到所有可能的回文串，再采用DP算法高效地解答多次查询请求。

You’ve got a string s = s1s2… s|s| of length |s|, consisting of lowercase English letters. There also are q queries, each query is described by two integers li, ri (1 ≤ li ≤ ri ≤ |s|). The answer to the query is the number of substrings of string s[li… ri], which are palindromes.

String s[l… r] = slsl + 1… sr (1 ≤ l ≤ r ≤ |s|) is a substring of string s = s1s2… 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 = t1t2… t|t| = t|t|t|t| - 1… t1.

Input
The first line contains string s (1 ≤ |s| ≤ 5000). The second line contains a single integer q (1 ≤ q ≤ 106) — the number of queries. Next q lines contain the queries. The i-th of these lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the description of the i-th query.
It is guaranteed that the given string consists only of lowercase English letters.

Output
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.

Examples

input
caaaba
5
1 1
1 4
2 3
4 6

output
1
7
3
4
2

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

题意就给你一个串，n次询问，每次询问一个区间内有多少个回文串。
区间型DP

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int SZ = 5010;

char S[SZ];
int get[SZ][SZ];
bool ha[SZ][SZ];

int main()
{
    int n, s, t;
    scanf("%s", S + 1);
    int len = strlen(S + 1);
    for(int i = 1; i <= len; i++)
        ha[i][i] = ha[i + 1][i] = get[i][i] = 1;
    for(int k = 2; k <= len; k++)
        for(int j, i = 1; i + k - 1 <= len; i++)
        {
            j = i + k - 1;
            ha[i][j] = (S[i] == S[j] && ha[i + 1][j - 1]);
            get[i][j] = get[i + 1][j] + get[i][j - 1] - get[i + 1][j - 1] + ha[i][j];
        }
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d%d", &s, &t);
        printf("%d\n", get[s][t]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/Loi-Vampire/p/6017047.html