poj1159 Palindrome

G - 回文串

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

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

题解：回文串：要求得到一个字符串变为回文串要插入字符的最少数

输入a字符串的长度，再输入a字符串；

将a字符串复制到b中（相当于将a字符串反置）

求出其与a的最长公共子序列len；

则min=strlen（a）-len

Eg.

A:       Ad3db

B:       bd3dA

最长公共子序列为d3d

所以len=3；

Min=n-len=2

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
short d[][];
int dp(char *s, int n)
{
    int i, j, k;
    for (i=n-;i>=;i--)
    {
        for (j=i+;j<n;j++)
        {
            if (s[i] == s[j])
            {
                d[i][j] = d[i+][j-];
            }
            else
            {
                d[i][j] = min(d[i][j-], d[i+][j]) + ;
            }
        }
    }
    return d[][n-];
}

int main( )
{
    char str[];
    int n;
    while (scanf("%d", &n) ==)
    {
        scanf("%s",str);
        printf("%d\n", dp(str, n));
    }
    return ;
}