poj1159

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 41570		Accepted: 14164

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

 

通过这样的结论可以和最长公共子串联系起来(未证明)：S和S' (注:S'是S的反串)的最长公共子串其实一定是回文的。 这样我们就可以借助lcs来解决该题，即 用s的长度减去lcs的值即可 。

 

#include <iostream>

#include <cstdio>

#include <cstring>



using namespace std;



const int N=5000;

short dp[N+1][N+1];

char str[N+1];

char r_str[N+1];

int len;



int main()

{

    while(scanf("%d",&len)!=EOF)

    {

        memset(dp,0,sizeof(dp));

        getchar();

        for(int i=1; i<=len; i++)

        {

            scanf("%c",&str[i]);

            r_str[len-i+1]=str[i];

        }

        //DP

        for(int i=1; i<=len; i++)

        {

            for(int j=1; j<=len; j++)

            {

                if(str[i]==r_str[j])

                {

                    dp[i][j]=dp[i-1][j-1]+1;

                }

                else

                {

                    dp[i][j]=dp[i-1][j]>dp[i][j-1]?dp[i-1][j]:dp[i][j-1];

                }

            }

        }

       printf("%d\n",len-dp[len][len]);

    }



    return 0;

}