POJ - 1159 Palindrome 最长公共子序列

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#dp #最长公共子序列

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本文介绍了一种求解给定字符串转换为回文所需的最小插入字符数的方法。通过计算字符串与其逆序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。示例代码展示了如何使用动态规划高效解决这一问题。

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 61919		Accepted: 21576

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 200

不难想就是求最长回文子序列长度，之前听别人讨论过最长回文子序长度的求法，是字符本身与它的逆序字符的最长公共子序列长度

例如 abbac

也就是求abbac 与cabba的最长公共子序列

写了一发超内存了，改了改才过了

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
typedef long long ll;
using namespace std;
int dp[3][5005];
int main()
{ 
	int i,j,n;
	char ch[5005],ch1[5005];
	scanf("%d",&n);
	getchar();
	scanf("%s",ch+1);
	for(i=1;i<=n;i++)
	ch1[n-i+1]=ch[i];
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
		{
			if(ch[i]==ch1[j])
			{
				dp[1][j]=dp[0][j-1]+1;
			}
			else
			dp[1][j]=max(dp[0][j],dp[1][j-1]);
		}
		for(j=1;j<=n;j++)
		dp[0][j]=dp[1][j];
	}
	printf("%d\n",n-dp[1][n]);
	return 0; 
 }