hdoj 1513 Palindrome (LCS 变形)

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5246    Accepted Submission(s): 1800

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

5Ab3bd

 

Sample Output

2

 

题目大意：所谓回文字符串，就是一个字符串，从左到右读和从右到左读是完全一样的，现在给你一个字符串，可在任意位置添加字符，最少再添加几个字符，可以使这个字符串成为回文字符串。

思路：原字符串与逆字符串的最长公共子序列

拓：这里用到一个滚动数组，可以节省空间，对时间没有作用，常用于动态规划，防止超内存

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[2][5050];//滚动数组 int main(){int n;char str[5050],s[5050];while(~scanf("%d",&n)){memset(dp,0,sizeof(dp));scanf("%s",str);//原序 for(int i=n-1;i>=0;i--)s[n-i-1]=str[i];//逆序 for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(str[i-1]==s[j-1])dp[i%2][j]=dp[(i-1)%2][j-1]+1;elsedp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]); }}printf("%d\n",n-dp[n%2][n]);}return 0;}