hdu1513 Palindrome

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
这题是最长公共子序列的一个变形，先求出所给字符串的逆字符串，然后求出最长公共自序，然后用n减去所求得的最大长度。
这里直接开dp[maxn][maxn]会超内存，所以用到滚动数组，因为每次只用到前2个，所以每次%2.
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 5060
int dp[2][maxn];
char s1[maxn],s2[maxn];
int main()
{
	int n,m,i,j,x,y;
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",s1+1);
		for(i=1;i<=n;i++){
			s2[i]=s1[n+1-i];
		}
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++){
			for(j=1;j<=n;j++){
				x=i%2;
				y=1-x;
				if(s1[i]==s2[j]){
					dp[x][j]=dp[y][j-1]+1;
				}
				else{
					dp[x][j]=max(dp[x][j-1],dp[y][j]);
				}
			}
		}
		printf("%d\n",n-dp[n%2][n]);
	}
	return 0;
}