POJ1159:动态规划

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 60290		Accepted: 20998

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

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN=;
unsigned short int dp[MAXN][MAXN];//dp[i][j]表示以第i个字母开头,第j个字母结尾的构成回文子串所需添加的字符串
char s[MAXN];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",s);
        for(int i=;i<n;i++)
        {
            dp[i][i]=;//有一个字符时
        }
        for(int i=;i<n;i++)
        {
            if(s[i-]==s[i])
                dp[i-][i]=;
            else
                dp[i-][i]=;
        }
        for(int k=;k<n;k++)
        {
            for(int i=,j=k;j<n;i++,j++)
            {
                if(s[i]==s[j])
                    dp[i][j]=dp[i+][j-];
                else
                    dp[i][j]=min(dp[i+][j],dp[i][j-])+;
            }
        }
        printf("%d\n",dp[][n-]);
    }
    return ;
}