PKU-1159 Palindrome (最長公共子序列LCS)

原題鏈接 http://poj.org/problem?id=1159

Palindrome

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 Code

/*
  最長公共子序列
  DP[i, j] = 0  if i == 0 or j == 0;
  DP[i, j] = DP[i - 1][j - 1] + 1;  if ch1[i] == ch2[j]
  DP[i, j] = max(DP[i - 1][j], DP[i][j - 1]);  if ch1[i] != ch2[j]

  滾動數組：
  dp[i][j]的值只依賴於dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1];
  dp[i%2][j]=max(dp[(i-1)%2][j], dp[i%2][j-1]);
  注意上面的取餘運算，重複利用這2行數組，數組好象在“滾動”一樣，所以叫滾動數組
*/

#include <iostream>
using namespace std;

#define Max 5002
#define max(a, b) (((a) > (b)) ? (a) : (b))

int main() {
	char input[Max], reversal[Max];
	int DP[2][Max];
	int i, j, len;

	while (scanf("%d", &len) !=EOF) {
		scanf("%s", input);
		for(i = 0; i < len; i++) {
			reversal[i] = input[len - 1 - i];
		}

		memset(DP, 0, sizeof(DP));

		for (i = 1; i <= len; i++) {
			for (j = 1; j <= len; j++) {
				if (input[i - 1] == reversal[j - 1]) {
					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", len - DP[len % 2][len]);
	}
	return 0;
}