hdoj 1513 Palindrome 【LCS】

Palindrome

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 17   Accepted Submission(s) : 2

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

 分析：

求给出数据的反序数据与原始数据的最长公共子序列的数与原长的差。

代码：

#include<cstdio>#include<algorithm>#include<cstring>#define max(a,b) (a>b?a:b)using namespace std;char s1[5010],s2[5010];int dp[2][5010];int main(){int n;while(scanf("%d",&n)!=EOF){scanf("%s",s1);int i,j;for(i=0;i<n;i++)s2[n-1-i]=s1[i];memset(dp,0,sizeof(dp));for(i=1;i<=n;i++)//因为数据要求很大，需要用滚动数组来记录。 {for(j=1;j<=n;j++){if(s1[i-1]==s2[j-1])dp[i%2][j]=dp[(i-1)%2][j-1]+1;elsedp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);}}printf("%d\n",n-dp[n%2][n]);}return 0;}