HDU - Palindrome(dp)

最新推荐文章于 2021-11-29 13:25:33 发布

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本文探讨如何通过最少字符插入操作将给定字符串转换为回文字符串，并提供了使用动态规划和滚动数组优化的解决方案。

Palindrome

Problem 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

这题其实LCS问题，将第一个序列反转，得到一个逆序列，然后，将这个问题转化为LCS问题，那么，可以使用dp来解决，但是，这个过程中，有一个细节要注意，由于题目的内存限制，没有优化的处理，将会出现MLE哦！这个时候可以使用滚动数组处理超内存的问题，那么，这个问题，主要的用到就是LCS + 滚动数组。下面是具体的AC代码：

#include <stdio.h>
#include <string.h>

#define MAX(x,y) ((x) > (y) ? (x) : (y))

int dp[2][5001];

int main(int argc , const char *argv)
{
    int n;
    char s1[5001],s2[5001];
    while(~scanf("%d",&n))
    {
    	scanf("%s",s1);
    	for(int i = 0; i < n; i++)
    	{
    		s2[n-i-1] = s1[i];
	    }
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                if(s1[i-1] == s2[j-1])
                {
                    dp[i%2][j] = dp[(i-1)%2][j-1] + 1;
                }
                else
                {
                    dp[i%2][j] = MAX(dp[(i-1)%2][j],dp[i%2][j-1]);
                }
            }
        }
        printf("%d\n",n - dp[n%2][n]);
    }
    return 0;
}