POJ-1159 Palindrome (区间dp)_区间dp poj

探讨了如何通过最少次数的字符插入将任意字符串转换为回文串的问题，使用动态规划算法实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 64675		Accepted: 22574

Description

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5
Ab3bd

Sample Output

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
short int dp[5001][5001];
char s[5001];
int main(){
	int n;
	scanf("%d", &n);
	scanf("%s", s);
	memset(dp, 0, sizeof(dp));
	for(int d = 1; d < n; ++d){
		for(int i = 0; i + d < n; ++i){
			if(s[i] == s[i + d]){
				dp[i][i + d] = dp[i + 1][i + d - 1];
			}
			else{
				dp[i][i + d] = 1 + min(dp[i][i + d - 1], dp[i + 1][i + d]);
			}
		}
	}
	printf("%d\n", dp[0][n - 1]);
}

/*
题意:
5000的字符串，问最小需要插入多少个字符可以构成一个回文串。

思路：
区间dp,dp[i][j]表示区间[i,j]需要最少需要插入多少个字符才可以构成一个回文串，
如果s[i] == s[j]，那么显然dp[i][j] = dp[i + 1][j - 1];如果不想等，那么就需要插入
一个字符，那么就是dp[i][j] = min(dp[i][j - 1], dp[i + 1][j]) + 1。注意用short int，
否则内存会超。
*/