hdoj 1513 Palindrome 【LCS】【滚动数组】

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4086    Accepted Submission(s): 1393

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
 
思路： 给你一个由n个字符组成的字符串，问最少再增加多少个字符就可以让其成为一个回文字符串，这道题用滚动数组做，因为dp[5000][5000]太大，超内存。
    求出原字符串的反转字符串，再求这两个字符串的最长公共子序列，最后用长度n减去最长公共子序列的长度即可。
    dp[i]可写成dp[i%2],dp[i-1]可写成dp[(i-1)%2].
 
代码：
 
#include<stdio.h>#include<string.h>#define max(a,b) (a>b?a:b)int dp[2][5100];char s1[5100],s2[5100];int main(){int n,i,j;while(scanf("%d",&n)!=EOF){scanf("%s",s1);strcpy(s2,s1);  //字符串复制； strrev(s2);  //字符串反转； memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){for(j=1;j<=n;j++){if(s1[i-1]==s2[j-1]){dp[i%2][j]=dp[(i-1)%2][j-1]+1;}else{dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);}}}printf("%d\n",n-dp[n%2][n]);}return 0;}