Palindrome HDU

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

大致题意：给你一个串，然你求出最少添加的字符数量使其变为一个回文串。

思路：将这个串翻转一下，然后求出两个串的最长公共子序列长度，那么所需添加的最少字符数量即原串长度减去该长度。
注意，这题需要使用滚动数组，否则会超内存。还有就是多组数据（坑）

代码如下

#include <iostream> #include <cmath>#include <algorithm>#include <cstring>#include <queue>#include <cstdio>#include <map>using namespace std; #define ll long long int char s2[5005];char s1[5005];int dp[2][5005];int main()  {        int n;    while(cin>>n>>s1)    {        int l1=strlen(s1);        for(int i=0;i<n;i++)        s2[n-1-i]=s1[i];        memset(dp,0,sizeof(dp));        for(int i=0;i<n;i++)        for(int j=0;j<n;j++)        {            int x=i+1,y=j+1;            if(s1[i]==s2[j])            dp[x%2][y]=dp[(x-1)%2][y-1]+1;            else            dp[x%2][y]=max(dp[(x-1)%2][y],dp[x%2][y-1]);        }        cout<<n-dp[n%2][n]<<endl;    }}