1159--Palindrome(dp:回文串变形2)

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. 

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.

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

5
Ab3bd

2

IOI 2000

另一种dp的方法，dp[l][i]，表示长度为l的以第i个字符开始的字符串需要多少操作。

对于长度为1的字符串，操作为0

长度为2的字符串，如果两个字符相同，操作为0，不同操作为1

长度为3的字符串，如果左右相同，操作为0，左右不同，对于a[1],a[2],a[3]来说 可以在前面增加a[3],或后面增加a[1]，那么就只需要判断剩余的两和字符需要的操作了。

长度为4的字符串，左右相同，那么需要求中间的两个字符，不同的话和长度为3的判断方式相同。

得到 长度为l开始为i的串可以由， 长度为l-1开始为i的，长度为l-1开始为i+1的，或者是长度为l-2，开始为i+1的变化得到。推出dp公式

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[3][5100] ;
char str[5100] ;
int main()
{
    int i , l , k1 , k2 , k3 , n ;
    while(scanf("%d", &n) !=EOF)
    {
        scanf("%s", str);
        for(i = n ; i >= 0 ; i--)
            str[i] = str[i-1] ;
        k1 = -1 ; k2 = 0 ; k3 = 1;
        memset(dp,0,sizeof(dp));
        for(l = 2 ; l <= n ; l++)
        {
            k3++ ;
            if(k3 == 3) k3 = 0 ;
            if(k3 == 0){ k2 = 2 ; k1 = 1 ; }
            else if( k3 == 1 ){ k2 = 0 ; k1 = 2 ; }
            else { k2 = 1 ; k1 = 0 ; }
            for(i = 1 ; i <= n-l+1 ; i++)
            {
                if( str[i] == str[i+l-1] )
                    dp[k3][i] = min( min(dp[k2][i]+1,dp[k2][i+1]+1),dp[k1][i+1] ) ;
                else
                    dp[k3][i] = min( dp[k2][i]+1 , dp[k2][i+1]+1);
            }
        }
        printf("%d\n", dp[k3][1]);
    }
    return 0;
}