HDU 1513 Palindrome （动态规划 & LCS）

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5509    Accepted Submission(s): 1872

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

5Ab3bd

 

Sample Output

2

 

Source

IOI 2000

 

【题目要求：】求输入的字符串能够组成回文字符串所需要插入的最少字符是多少？

【思路：】把输入的字符串进行倒置，问题即可转化成求两个字符串最长公共子序列的问题，最后用字符串长度减去最长公共子序列即可得要求结果。因为5000得二维数组会超出内存，所以采用滚动数组。

最长公共子序列问题讲解：http://blog.csdn.net/i1020/article/details/54918665

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;char a[5005],b[5005];int dp[2][5005];int main(){int n;while(scanf("%d",&n) != EOF){getchar();for(int i = 0; i < n; i ++){cin >> a[i];b[n - 1 - i] = a[i];}memset(dp,0,sizeof(dp));for(int i = 1; i <= n; i ++)for(int j = 1; j <= n; j ++){if(a[i - 1] == b[j - 1])dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1;elsedp[i % 2][j] = max(dp[i % 2][j - 1],dp[(i - 1) % 2][j]);}printf("%d\n",n - dp[n % 2][n]);}return 0;}