杭电-1513Palindrome（LCS+滚动数组）

最新推荐文章于 2016-08-12 21:04:35 发布

原创最新推荐文章于 2016-08-12 21:04:35 发布 · 525 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#杭电 #1513 #Palindrome #LCS+滚动数组

HDU—ACM 同时被 2 个专栏收录

82 篇文章

订阅专栏

最长公共子序列

7 篇文章

订阅专栏

本文介绍了一种算法，用于确定将任意字符串转换为回文所需的最少字符插入次数。通过使用动态规划方法，并采取滚动数组优化内存使用，该算法能够高效解决此问题。

Palindrome

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4344 Accepted Submission(s): 1482

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

明显的LCS，但是要用到二位数组dp[5000][5000]会超内存，联系到每次的计算只与上次有关，所以可以用到滚动数组！

#include<cstdio>
#include<cstring>
#include<algorithm> 
using namespace std;
int dp[2][5500];
char s1[5500],s2[5500];
int main()
{
	int m,n,i,j;
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",s1);
		int t=0;
		for(i=n-1;i>=0;i--)
			s2[t++]=s1[i];
		s2[n]='\0';
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				int x=i%2;
				int y=1-x;
				if(s1[i-1]==s2[j-1])
				dp[x][j]=dp[y][j-1]+1;
				else
				dp[x][j]=max(dp[y][j],dp[x][j-1]);
			}
		}
		printf("%d\n",n-(dp[n%2][n]));
	}
	return 0;
}