POJ 1159 Palindrome(DP LCS)

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：求插入最少的字符使得串为palindeome（回文串）。

遇到问题要剖析其原型，此题就是正序和逆序的LCS，结果来n-commenlength.

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
short dp[5001][5001];
char a[5005],b[5005];
int main()
{
    int n,m,x,k,i,j;
    int cla;
    while(~scanf("%d",&n))
    {
        int s=0;
        dp[0][0]=0;<span id="transmark"></span>
        scanf("%s",a);
        for(i=n-1;i>=0;i--)
        b[s++]=a[i];
        b[s]='\0';
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(a[i-1]==b[j-1])
                {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
            }
        }
        cout<<n-dp[n][n]<<endl;
    }
    return 0;
}

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