HDOJ 1513 Palindrome【回文串&&滚动数组】

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5165    Accepted Submission(s): 1761

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

 

这里主要有一个回文串的问题。一个序列的 最长回文子序列就是它和自己逆序的最长公共子序列（LCS）。

然后因为5000太大了，开不了二维数组，所以我们可以用滚动数组，就是奇数偶数交叉记录。因为这个题只需要最后结果，所以中间过程被省去也没有问题。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[2][5010];
char ch1[5010],ch2[5010];
int main()
{
	int i,j,n;
	while(~scanf("%d",&n))
	{
		memset(dp,0,sizeof(dp));
		scanf("%s",ch1);
		for(i=0;i<n;i++)
			ch2[i]=ch1[n-i-1];
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				if(ch1[i-1]==ch2[j-1])
					dp[i&1][j]=dp[(i-1)&1][j-1]+1;//用了位运算，也不知道节省的时间多不多=。=
				else
					dp[i&1][j]=max(dp[(i-1)&1][j],dp[i&1][j-1]);
			}
		}
		printf("%d\n",n-dp[n&1][n]);
	}
	return 0;
}