【poj 1159】 Palindrome DP(类最长公共子序列)+滚动数组_dp最长公共子序列滚动数组-优快云博客

本文介绍了一个算法问题，即如何通过最少的字符插入步骤将任意字符串转化为回文串。文章提供了完整的实现代码，并采用动态规划方法计算原字符串与其逆序字符串之间的最长公共子序列。

Palindrome
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 58492 Accepted: 20318

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

题目链接：http://poj.org/problem?id=1159；

题意：求一个序列改造构成回文串最小步数；

思路：计算串与其逆串类最长公共子序列； ans=len-maxcom
*因为500*5000，dp[i][j]只与dp[i-1][]有关用滚动数组优化

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int n;
char s1[5005];
char s2[5005];
int dp[2][5005];
int main()
{
    scanf("%d",&n);
    scanf("%s",s1+1);
    for(int i=1;i<=n;i++)
    s2[i]=s1[n-i+1];
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)
    {   
        if(s1[i]==s2[j]) 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]);
}