杭电-1513Palindrome（LCS+滚动数组）

拼搏现实的明天。 2022-08-10 05:46 147阅读 0赞

# Palindrome #

**Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)**

Total Submission(s): 4344 Accepted Submission(s): 1482

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

明显的LCS，但是要用到二位数组dp\[5000\]\[5000\]会超内存，联系到每次的计算只与上次有关，所以可以用到滚动数组！

#include<cstdio>
    #include<cstring>
    #include<algorithm> 
    using namespace std;
    int dp[2][5500];
    char s1[5500],s2[5500];
    int main()
    {
    	int m,n,i,j;
    	while(scanf("%d",&n)!=EOF)
    	{
    		scanf("%s",s1);
    		int t=0;
    		for(i=n-1;i>=0;i--)
    			s2[t++]=s1[i];
    		s2[n]='\0';
    		memset(dp,0,sizeof(dp));
    		for(i=1;i<=n;i++)
    		{
    			for(j=1;j<=n;j++)
    			{
    				int x=i%2;
    				int y=1-x;
    				if(s1[i-1]==s2[j-1])
    				dp[x][j]=dp[y][j-1]+1;
    				else
    				dp[x][j]=max(dp[y][j],dp[x][j-1]);
    			}
    		}
    		printf("%d\n",n-(dp[n%2][n]));
    	}
    	return 0;
    }