poj 1159 -- Palindrome

Palindrome

 

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 53250		Accepted: 18396

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

---

题目连接：Palindrome

思路：简单dp。头尾来一次最长公共子序列。用长度n-最长公共子序列即为答案。

 1 /*======================================================================
 2  *           Author :   kevin
 3  *         Filename :   Palindrome.cpp
 4  *       Creat time :   2014-09-25 10:31
 5  *      Description :
 6 ========================================================================*/
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstdio>
10 #include <cstring>
11 #include <queue>
12 #include <cmath>
13 #define clr(a,b) memset(a,b,sizeof(a))
14 #define M 5005
15 using namespace std;
16 char str1[M];
17 short int dp[M][M];
18 int main(int argc,char *argv[])
19 {
20     int n;
21     while(scanf("%d",&n) != EOF){
22         scanf("%s",str1);
23         for(int i = 1; i <= n; i++){
24             for(int j = 1; j <= n; j++){
25                 if(str1[i-1] == str1[n-j]){
26                     dp[i][j] = dp[i-1][j-1] + 1;
27                 }
28                 else{
29                     dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
30                 }
31             }
32         }
33         printf("%d\n",n - dp[n][n]);
34     }
35     return 0;
36 }

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