HDU 1513 Palindrome【LCS+滚动数组】

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3423    Accepted Submission(s): 1181

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
   
   
   
   

 

题意：给出一个字符串，问要将这个字符串变成回文串要添加最少几个字符

思路：将该字符串与其反转求一次LCS，然后所求就是n减去最长公共子串的长度，但是要注意这里字符串最长有5000，dp数组二维都开5000的话就会超内存，这里就用到了滚动数组，因为在LCS的计算中，i的变化只相差1，所以可以通过对2取余来进行滚动。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdlib>
using namespace std;

int dp[2][5500],n;
char s1[5500], s2[5500];

void LCS(){
	int i,j;
	memset(dp, 0, sizeof(dp));
	for(i = 1; i <= n; ++i)
		for(j = 1; j <= n; ++j){
			if(s1[i-1] == s2[j-1])
				dp[i%2][j] = dp[(i-1)%2][j-1]+1;
			else dp[i%2][j] = max(dp[i%2][j-1],dp[(i-1)%2][j]);
		}
}

int main (){
	while(scanf("%d", &n)!=EOF){
		scanf("%s", s1);
		strcpy(s2,s1);
		reverse(s2,s2+n);
		//cout<<s2<<endl;
		LCS();
		printf("%d\n", n-dp[n%2][n]);
	}
	return 0;
}