Palindrome
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 57568 Accepted: 19962
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

題意：
有一個字串。要求問最少新增幾個數使得它是一個迴文串
題解：
觀察後可以發現，這個就是將字串倒置後與原串求最大公共子列的長度，然後n-公共子列長度就是需要新增字元的個數。馬上就寫了一個最長子序列的dp，結果交上去一直超記憶體。 。後來才反應過來用滾動陣列優化記憶體。這個滾動陣列有兩行，old表示之前求得的那一行結果，now表示現在正在遞推的一行。然後操作完一行後對old和now進行異或操作。就把old和now的值換了一下。這樣陣列就是dp[2][5010]，就不會超記憶體了。PS：這題用short int也能過。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define F(i,a,b) for(int i = a;i<=b;i++)
#define FI(i,a,b) for(int i = a;i>=b;i--)
#define SIZE 5000+10

using namespace std;

int dp[2][SIZE];
char a[SIZE];

int solve(int n){
    int old = 0,now = 1;
    memset(dp,0,sizeof(dp));
    F(i,1,n){
        FI(y,n,1){
            int j = n+1-y;
            if(a[i]==a[y]){
                dp[now][j] = dp[old][j-1] + 1;
            }else{
                dp[now][j] = max(dp[old][j],dp[now][j-1]);
            }
        }
        now^=1;
        old^=1;
    }
    return dp[old][n];
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        getchar();
        F(i,1,n){
            char c = getchar();
            a[i] = c;
        }
        getchar();
        printf("%d\n",n-solve(n));
    }


    return 0;
}