CodeForces 245H Queries for Number of Palindromes (区间DP)

最新推荐文章于 2021-01-26 16:00:46 发布

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本文介绍了一种高效计算字符串中特定区间内回文子串数量的算法，并提供了完整的C++实现代码。该算法通过预处理得到所有可能的回文子串信息，能够在O(1)时间内回答任意区间的回文子串数量。

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 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_ls_l + 1... s_r (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 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.

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.

Sample test(s)

input

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

output

首先ispal[i][j]表示字符串从 i 到 j 这一段是否为回文，若是，则为1。不难得到初始化时ispal[i][i] = 1。那么，对于长度为len的字符串，判断它是否为回文，则有ispal[i][i+len-1] = ispal[i+1][i+len-2]&&str[i]==str[i+len-1];（同时ispal[i][i-1]要也要初始化为1，因为当len=2时，ispal[i+1][i+len-2] = ispal[i+1][i]，需要拿来计算）

dp[i][j]表示从 i 到 j 包含的回文串的个数，同样的，初始化dp[i][i] = 1。计数时有：

dp[i][i+len-1] = dp[i][i+len-2]+dp[i+1][i+len-1]-dp[i+1][i+len-2]+ispal[i][i+len-1];然后O(n^2)把每个区间的结果计算出来，对于询问s到e，直接输出dp[s][e]即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define SIZE 5001

using namespace std;

int ispal[SIZE][SIZE];
int dp[SIZE][SIZE];
char str[SIZE];
int Q;

int main()
{
    while(~scanf("%s",str+1))
    {
        int len = (int)strlen(str+1);
        for(int i=1; i<=len; i++)
        {
            dp[i][i] = ispal[i][i] = 1;
            ispal[i][i-1] = 1;
        }
        for(int i=2; i<=len; i++)
        {
            for(int s=1; s+i-1 <=len; s++)
            {
                ispal[s][s+i-1] = ispal[s+1][s+i-2]&&str[s]==str[s+i-1];
                dp[s][s+i-1] = dp[s][s+i-2]+dp[s+1][s+i-1]-dp[s+1][s+i-2]+ispal[s][s+i-1];
            }
        }
        scanf("%d",&Q);
        int s,e;
        while(Q--)
        {
            scanf("%d%d",&s,&e);
            printf("%d\n",dp[s][e]);
        }
    }
    return 0;
}