hdoj 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): 3918    Accepted Submission(s): 1340

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：最少向字符串中添加几个字符可以使该字符串变成回文字符串

题解：将字符串反转，跟原字符串使用LCS，算出最长公共子序列长度，拿总长度减去最长公共子序列长度就是答案

滚动数组详解：http://blog.csdn.net/niushuai666/article/details/6677982

#include<stdio.h>
#include<string.h>
#define MAX 5010
#define max(x,y)(x>y?x:y)
char s1[MAX],s2[MAX];
int dp[2][MAX];
int main()
{
	 int i,j;
	 int t;
	 while(scanf("%d",&t)!=EOF)
	 {
	 	scanf("%s",s1);
	 	for(i=t-1,j=0;i>=0;i--)
	 	s2[j++]=s1[i];
	 	memset(dp,0,sizeof(dp));
	 	for(i=1;i<=t;i++)
	 	{
	 		for(j=1;j<=t;j++)
	 		{
	 			dp[0][j]=dp[1][j];
	 		    dp[1][j]=dp[2][j];
	 			if(s1[i-1]==s2[j-1])
	 			dp[1][j]=dp[0][j-1]+1;
	 			else
	 			dp[1][j]=max(dp[1][j-1],dp[0][j]);
			}
		}
		printf("%d
",t-dp[1][t]);
	 }
	return 0;
}