hdu1513——Palindrome

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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。通过动态规划方法，该算法能够高效地解决这一问题，并提供了一个实际的C++代码示例。

Palindrome

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3281 Accepted Submission(s): 1136

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

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LCS，加个滚动数组

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int dp[2][5010];
char a[5010], b[5010];

int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		scanf("%s", a);
		for (int i = 0; i < n; i++)
		{
			b[i] = a[n - i - 1];
		}
		b[n] = '\0';
		for (int i = 0; i <= n; i++)
		{
			dp[0][i] = 0;
			dp[1][i] = 0;
		}
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (a[i - 1] == b[j - 1])
				{
					dp[i % 2][j] = dp[1 - i % 2][j - 1] + 1;
				}	
				else
				{
					dp[i % 2][j] = max(dp[1 - i % 2][j], dp[i % 2][j - 1]);
				}
			}
		}
		printf("%d\n", n - dp[n % 2][n]);
	}
	return 0;
}