poj1159-Palindrome(메모 문자열을 구성하는 최소 문자 수, dp)

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
일반적인 방법은 이 직렬과 그 역렬의 최대 공통 서열 길이를 구한 다음에 직렬의 길이로 이 길이를 줄이는 것이다. 또 하나는 dp[i][j]로 i에서 j까지 추가해야 하는 최소 문자수인 if(a[i]=a[j])dp[i][j]=dp[i]=dp[i+1][j-1]elsedp[i][j]=min(dp[i+1][j], dp[j][j][j][j]]

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MAXN 5010
using namespace std;
char a[MAXN];
short dp[MAXN][MAXN]; //  ，short  ac
int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        scanf("%s",a+1);
        for(i=n; i>=1; --i)
        {
            dp[i][i]=0;
            for(j=i+1; j<=n; ++j)
            {
                if(a[i]==a[j])
                    dp[i][j]=dp[i+1][j-1];
                else
                    dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;
            }
        }
        printf("%d
",dp[1][n]);
    }
    return 0;
}