POJ 1159 Palindrome（回文字符串）

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 54047		Accepted: 18672

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

大致题意：给定一个字符串，问最少插入多少字符，使该字符变成回文字符串。

思路：设原字符串序列为X，逆序列为Y，则最少需要补充的字母数=X的长度-X和Y的最长公共子序列的长度。这道题有意思的地方不是发现这个公式，而是对空间的压缩处理。出具范围达到了5000，若开静态数组用int肯定会超，当时用short就会过了。最好的办法就是用滚动数组。求LCS的状态转移方程为：d[i][j]=d[i-1][j]+d[i][j-1]；上面的d[i][j]只依赖于d[i-1][j]，d[i][j-1]；运用滚动数组d[i%2][j]=d[(i-1)%2][j]+d[i%2][j-1]；滚动数组d[i%2][j]=d[(i-1)%2][j]+d[i%2][j-1];滚动数组实际是一种节省空间的方法，在时间上没有优势。

#include<iostream>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<sstream>
#include<vector>
using namespace std;
int dp[2][5005];
int main()
{
    string s1,s2;
    int i,n,j;
    while(cin>>n)
    {
        cin>>s1;
        s2=s1;
        reverse(s1.begin(),s1.end());
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);
                if(s1[i-1]==s2[j-1])
                {
                    int temp=dp[(i-1)%2][j-1]+1;
                    dp[i%2][j]=max(dp[i%2][j],temp);
                }
            }
        }
        cout<<n-dp[n%2][n]<<endl;
    }
    return 0;
}