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

中文题意

以下为官方英文题意（课堂讲解时请要求学生能用自己的话复述条件与目标）。

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?

Note（官方）： 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 $0$.

输入格式（中文）

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.

输出格式（中文）

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