poj 1159 && hdu 1513 Palindrome(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

5Ab3bd

Sample Output

2

solution:

问在任意位置填最少的字符，使得这个串为回文串。设原序列S的逆序列为S’，则最少需要补充的字母数=原序列S的长度-S和S’的最长公共子串长度

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[2][5500],n;char s1[5500], s2[5500];int main(){while (~scanf("%d", &n)){scanf("%s", s1+1);for (int i = 1; i <= n; i++)s2[i] = s1[n - i + 1];memset(dp, 0, sizeof(dp));for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (s1[i] == s2[j])dp[i%2][j] = dp[(i - 1)%2][j - 1] + 1;else dp[i%2][j] = max(dp[(i - 1)%2][j], dp[i%2][j - 1]);printf("%d\n", n - dp[n%2][n]);}return 0;}