北大POJ1159 Palindrome（动态规划求最长公共子序列）

                                                       Palindrome
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

题目大意：给出一个字符串，计算出最少添加几个字符使它变成回文字符串（从前向后和从后向前读字符串是一样的）

解题思路：求出字符串和它的倒置字符串的字长公共子序列，用总长度减去最长公共子序列的长度即可得到答案，因为在最长公共子序列的部分不需要再添加字符，而在不是最长公共子序列的位置上插入和另外一个字符串数组相同的字符即可。

AC代码：

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

char str1[5010],str2[5010];

short f[5010][5010];  //当定义成int类型时会超内存，定义成short时，内存：49340K，时间：1032MS

int max(int a,int b)  //求最大值

{

    if(a>=b)

    return a;

    else

    return b;

}

void exchange(int n) //将输入的字符串数组倒置

{

     int i,j;

     for(i=1;i<=n;i++)

     {

     scanf("%c",&str1[i]);

     }

     for(i=n,j=1;i>=1;i--,j++)

     {

     str2[j]=str1[i];

     }

}

int  dp(int n)   //动态规划求最长公共子序列，核心代码

{   

     int i,j;

     memset(f,0,sizeof(f));

     for(i=1;i<=n;i++)

     {

     for(j=1;j<=n;j++)

     {

     if(str1[i]==str2[j])

     f[i][j]=f[i-1][j-1]+1;

     else if(str1[i]!=str2[j])

     f[i][j]=max(f[i-1][j],f[i][j-1]);

     }

     }

     return f

;

}

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

    getchar();

    exchange(n);

    printf("%d\n",n-dp(n));  

    }

    return 0;

}