poj-1159-Palindrome【LCS】【滚动数组】

最新推荐文章于 2021-02-13 15:57:03 发布

Somethingwll

最新推荐文章于 2021-02-13 15:57:03 发布

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分类专栏：大菜鸟小历程 LCS 滚动数组

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本文介绍了一个算法问题的解决方案，通过计算字符串转换为回文串所需的最少字符插入次数。利用动态规划方法，计算原始字符串与其反转字符串之间的最长公共子序列长度，进而得出所需插入的字符数量。

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题目链接：点击打开链接

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 60656		Accepted: 21130

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

记着这个结论，需要加的字符的个数 = 原来字符串的长度 - 原来字符串和逆字符串的最长公共子序列的长度。因为空间有限所以要用

滚动数组即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
char str1[5010],str2[5010];
int dp[2][5010];
int main()
{
	while(~scanf("%d",&n))
	{
		scanf("%s",str1);
		for(int i=0;i<n;i++)
			str2[i]=str1[n-i-1];
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(str1[i-1]==str2[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;
}