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

H. Queries for Number of Palindromes

time limit per test
5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Sample test(s)

input

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

output

1
7
3
4
2

 

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;
}