Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4062    Accepted Submission(s): 1384

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

Source

IOI 2000

Recommend

linle   |   We have carefully selected several similar problems for you:  1505 1506 1074 1003 1025

RE:怎么把一个字符串变为回文串？ → → 反转后求LCS； 为了不爆内存用滚动数组，吐个槽： 这题WA到恶心。

　　　滚动数组：http://blog.csdn.net/insistgogo/article/details/8581215

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 char a[5050], b[5050];
 6 int dp[2][5050];
 7 int main()
 8 {
 9     int n;
10     while(~scanf("%d", &n))
11     {
12         scanf("%s", b);
13         strcpy(a, b);
14         strrev(b);
15         memset(dp, 0, sizeof(dp));           //+F+++U++++K++
16         for(int i = 1; i <= n; i++)
17             for(int j = 1; j <= n; j++)
18             {
19                 if(a[i-1] == b[j-1])
20                     dp[i%2][j] = dp[(i-1)%2][j-1] + 1;
21                 else
22                     dp[i%2][j] = max(dp[(i-1)%2][j], dp[i%2][j-1]);
23             }
24             printf("%d\n", n - dp[n%2][n]);
25     }
26     return 0;
27 }