CF1509C · The Sports Festival

• 难度：1800
• 标签：dp、greedy
• 链接：https://codeforces.com/problemset/problem/1509/C
• 时间限制：1 second　内存限制：256 megabytes
• 出现位置：假期自习包/F-区间DP入门

英文原题面

Statement

The student council is preparing for the relay race at the sports festival. The council consists of $n$ members. They will run one after the other in the race, the speed of member $i$ is $s_i$. The discrepancy $d_i$ of the $i$-th stage is the difference between the maximum and the minimum running speed among the first $i$ members who ran. Formally, if $a_i$ denotes the speed of the $i$-th member who participated in the race, then $d_i = \max(a_1, a_2, \dots, a_i) - \min(a_1, a_2, \dots, a_i)$. You want to minimize the sum of the discrepancies $d_1 + d_2 + \dots + d_n$. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?

Input

The first line contains a single integer $n$ ($1 \le n \le 2000$)  — the number of members of the student council. The second line contains $n$ integers $s_1, s_2, \dots, s_n$ ($1 \le s_i \le 10^9$)  – the running speeds of the members.

Output

Print a single integer  — the minimum possible value of $d_1 + d_2 + \dots + d_n$ after choosing the order of the members.

样例

样例 1

输入：

3
3 1 2

输出：

3

样例 2

输入：

1
5

输出：

0

样例 3

输入：

6
1 6 3 3 6 3

输出：

11

样例 4

输入：

6
104 943872923 6589 889921234 1000000000 69

输出：

2833800505

样例解释（英文原文）

In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus $a_1 = 2$, $a_2 = 3$, and $a_3 = 1$. We have:

• $d_1 = \max(2) - \min(2) = 2 - 2 = 0$.
• $d_2 = \max(2, 3) - \min(2, 3) = 3 - 2 = 1$.
• $d_3 = \max(2, 3, 1) - \min(2, 3, 1) = 3 - 1 = 2$.

The resulting sum is $d_1 + d_2 + d_3 = 0 + 1 + 2 = 3$. It can be shown that it is impossible to achieve a smaller value. In the second test case, the only possible rearrangement gives $d_1 = 0$, so the minimum possible result is $0$.