HDU 1513 Palindrome （滚动数组lcs，水）

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Palindrome Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6427    Accepted Submission(s): 2123 Problem 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   Source IOI 2000   Recommend linle   Statistic | Submit | Discuss | Note

水题，不多说了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 5e3 + 5;
int dp[2][maxn];
char s1[maxn], s2[maxn];
int main()
{
int n;
while(~scanf("%d", &n))
{
memset(dp, 0, sizeof(dp));
scanf("%s", s1+1);
int index = n, cur = 0;
for(int i = 1; i <= n; i++)
s2[index--] = s1[i];
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if(s1[i] == s2[j])
dp[cur][j] = dp[cur^1][j-1] + 1;
else
dp[cur][j] = max(dp[cur^1][j], dp[cur][j-1]);
}
cur ^= 1;
}
printf("%d\n", n - dp[cur^1]
);
}
return 0;
}