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): 4113 Accepted Submission(s): 1405

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

题意：向一个字符串中添加其本身的字符，最少添加几个能成为回文串。

解题思路：求出字符串的逆序串，找出逆序串与本串的最长公共子序列，即用原串的长度减去子串的长度，可得出需要添加的最小字符的个数。

注意，本题的字符串长度最大可达到5000，dp数组[5000][5000]会爆内存，所以用滚动数组dp[2][5050]。

代码如下：

<span style="font-size:12px;">#include<cstdio>
#include<cstring>
char str[5050],s[5050];
int dp[2][5050];

int max(int a,int b)
{
    return a>b?a:b;
}

int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        scanf("%s",str);
        for(i=0;i<n;++i)
            s[i]=str[n-1-i];
        for(i=1;i<=n;++i)
        {
            for(j=1;j<=n;++j)
            {
                if(str[i-1]==s[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]
);
    }
    return 0;
} </span>