1159--Palindrome(dp:回文串变形2)

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 53431		Accepted: 18454

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

Source

IOI 2000

还有一种dp的方法，dp[l][i]，表示长度为l的以第i个字符開始的字符串须要多少操作。

对于长度为1的字符串，操作为0

长度为2的字符串，假设两个字符同样，操作为0，不同操作为1

长度为3的字符串。假设左右同样，操作为0。左右不同，对于a[1],a[2],a[3]来说 能够在前面添加a[3],或后面添加a[1]，那么就仅仅须要推断剩余的两和字符须要的操作了。

长度为4的字符串，左右同样，那么须要求中间的两个字符，不同的话和长度为3的推断方式同样。

得到 长度为l開始为i的串能够由， 长度为l-1開始为i的，长度为l-1開始为i+1的。或者是长度为l-2，開始为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;
}