HDOJ题目1513 Palindrome（LCS，滚动数组）

最新推荐文章于 2016-08-13 10:50:15 发布

Jogging_Clown

最新推荐文章于 2016-08-13 10:50:15 发布

阅读量503

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划

本文链接：https://blog.youkuaiyun.com/yu_ch_sh/article/details/41594265

动态规划专栏收录该内容

194 篇文章

订阅专栏

本文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，并提供了一种利用动态规划解决该问题的有效方法。具体而言，文章介绍了一个基于最长公共子序列（LCS）的算法实现，该算法采用滚动数组来降低空间复杂度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Palindrome

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3252 Accepted Submission(s): 1126

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

Source

IOI 2000

Recommend

linle | We have carefully selected several similar problems for you: 1074 1510 1501 1500 1422

思路：正反串取最长公共子序列，然后总长度n减去就行，注意字符串长度是最长是5000，用dp[5001][5001]会超内存，所以使用

滚动数组

ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int dp[2][5005],n;
void lcs(char *s1,char *s2)
{
	int i,j;
	memset(dp,0,sizeof(dp));
	for(i=1;i<=n;i++)
	{
		for(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]);
		}
	}
}
int main()
{
	//int n;
	while(scanf("%d",&n)!=EOF)
	{
		char s1[5005],s2[5005];
		int i,j=n;
		scanf("%s",s1);
		s2[j--]='\0';
		for(i=0;i<n;i++)
		{
			s2[j--]=s1[i];
		}
		lcs(s1,s2);
		printf("%d\n",n-dp[n%2][n]);
	}
}