Palindrome POJ 1159 【LCS】

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

   
   
   
   

    
    
    
    2

   
   
   
   

 

/*
题意： 
最少插入多少个数，使某字符串回文 
res=原字符串的长度 - 原字符串和源字符串的逆字符串的最长公共子序列的长度 
另外二维数组最多只能定义到1000，这里可以采用滚动数组，即当用完了一次再给它赋上一个我们需要的值，也就是循环利用 
*/


#include<stdio.h>
#include<string.h>
#define max(a,b) (a)>(b)?(a):(b)
char s1[5010],s2[5010];
int dp[3][5010];
int main()
{
    int n;
    int i,j;
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        scanf("%s",s1);
        j=0;
        for(i=n-1;i>=0;--i)
            s2[j++]=s1[i];
        for(i=1;i<=n;++i)
        {
            for(j=1;j<=n;++j)
            {
                if(s1[i-1]==s2[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;
}