LCS最长公共子序列~dp学习~4

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1513

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4532    Accepted Submission(s): 1547

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

 

Source

IOI 2000

题意：给出一个字符串，问要将这个字符串变成回文串要添加最少几个字符

思路：将该字符串与其反转求一次LCS，然后所求就是n减去最长公共子串的长度，但是要注意这里字符串最长有5000，dp数组二维都开5000的话就会超内存，这里就用到了滚动数组，因为在LCS的计算中，i的变化只相差1，所以可以通过对2取余来进行滚动

LCS： 求连个串s1,s2的最长公共自序列：dp[i][j] 表示扫描到第一个串的第i个位置第二个串的第j个位置的最长公共子序列。 当s1[i]==s2[j]时，dp[i][j] = d[i-1][j-1]+1;

当s1[i]!=s2[j]时，dp[i][j] = max(dp[i-1][j],dp[i][j-1]);

空间优化： 通过上面的转移方程可以看出来dp[i][j]的状态之和上一列和当前列有关系，所以可以通过二维滚动数组的形式来储存，通过i的奇偶来控制

下面是代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 5005;
char s1[N],s2[N];
int dp[2][N];
int n;

void LCS()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    for(i = 1; i<=n; i++)
    {
        for(j = 1; j<=n; j++)
        {
            int x = i%2;
            int y = 1-x;
            if(s1[i-1]==s2[j-1])
                dp[x][j] = dp[y][j-1]+1;
            else 
                dp[x][j] = max(dp[y][j],dp[x][j-1]);
        }
    }
}
int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        getchar();
        scanf("%s",s1);
        for(i = 0; i < n; i++)
            s2[i] = s1[n-1-i];
        //s2[i] = '';
        LCS();
        printf("%d
",n-dp[n%2][n]);
    }
    return 0;
}