POJ 1159 - Palindrome (LCS， 滚动数组)

Palindrome

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】 给你一个长度为n的字符串，问最少再添多少字符能组成一个回文串；【分析】原字符串：Ab3bd翻转后串：db3ba二者有重复子串b3b,若想构成回文串，必须要再添加除重复子串外的其他字符。如：Adb3bdA

下面的问题就是求原字符串与翻转后串的最长公共子串，即LCS问题；

【LCS问题】标记s1,s2字符位置变量i,j，令dp[i][j]为字符串s1[1~i],s2[1~j]的最长公共子串的长度;可知状态转移方程如下：dp[i][j] = s1[i] == s2[j] ? dp[i-1][j-1] : max(dp[i-1][j], dp[i][j-1]);

【注意】对于本题，n的范围是[3,5000]，若直接开5000*5000的二维数组会内存超限（当然听说用short int会AC飘过）；

【滚动数组】滚动数组的作用在于优化空间。主要应用在递推或动态规划中（如01背包问题）。因为DP题目是一个自底向上的扩展过程，我们常常需要用到的是连续的解，前面的解往往可以舍去。所以用滚动数组优化是很有效的。利用滚动数组的话在n很大的情况下可以达到压缩存储的作用。

例如本题，dp[i][j]的值仅仅取决于dp[i-1][j-1], dp[i][j-1], dp[i-1][j];再直白地说，只需要保留下i-1时的状态，就可以求出i时的状态；所以dp完全可以只开一个2*5000的数组求解；或许有人问j为什么不能也开成2? 这很好说明，因为j是随i不断循环的，i增加一个j全部循环一次，所以i在不断变化时需要不断j全部的信息，我们完全也可以令i随j不断变化，这样仅仅改变成5000*2，其他完全一样；

【代码】

 1 /*LCS*/
 2
 3 #include<iostream>
 4 #include<cstdio>
 5 #include<cstdlib>
 6 #include<cstring>
 7 using namespace std;
 8 const int maxn = 5010;
 9 char s1[maxn], s2[maxn];
10 int n;
11 int dp[3][maxn];
12
13 void LCS()
14 {
15     memset(dp, 0, sizeof(dp));
16
17     for(int i = 1; i <= n; i++)
18     {
19         for(int j = 1; j <= n; j++)
20         {
21             //cout << s1[i] << " " << s2[j] << endl;
22             if(s1[i] == s2[j])
23                 dp[i%2][j] = dp[(i-1)%2][j-1]+1;
24             else
25                 dp[i%2][j] = max(dp[(i-1)%2][j], dp[i%2][j-1]);
26         }
27     }
28     //cout << dp[n%2][n] << endl;
29     printf("%d\n", n-dp[n%2][n]);
30
31
32 }
33
34 int main()
35 {
36     while(~scanf("%d", &n))
37     {
38         scanf("%s", s1+1);
39
40         for(int i = 0; i < n; i++)
41             s2[i+1] = s1[n-i];
42
43         LCS();
44
45     }
46     return 0;
47 }