- 题目提供者 FarmerJohn2
- 评测方式 云端评测
- 标签 USACO 2013
- 难度 提高+/省选-
- 时空限制 1000ms / 128MB
Farmer John has forgotten to repair a hole in the fence on his farm, and his N cows (1 <= N <= 1,000) have escaped and gone on a rampage! Each minute a cow is outside the fence, she causes one dollar worth of damage. FJ must visit each cow to install a halter that will calm the cow and stop the damage.
Fortunately, the cows are positioned at distinct locations along a straight line on a road outside the farm. FJ knows the location P_i of each cow i (-500,000 <= P_i <= 500,000, P_i != 0) relative to the gate (position 0) where FJ starts.
FJ moves at one unit of distance per minute and can install a halter instantly. Please determine the order that FJ should visit the cows so he can minimize the total cost of the damage; you should compute the minimum total damage cost in this case.
农夫约翰的牧场围栏上出现了一个洞，有N(1 <= N <= 1,000)只牛从这个洞逃出了牧场。这些出逃的奶牛很狂躁，他们在外面到处搞破坏，每分钟每头牛都会给约翰带来1美元的损失。约翰必须用缰绳套住所有的牛，以停止他们搞破坏。
幸运的是，奶牛们都在牧场外一条笔直的公路上，牧场的大门恰好位于公里的0点处。约翰知道每头牛距离牧场大门的距离P_i(-500,000 <= P_i <= 500,000, P_i != 0)
* Line 1: The number of cows, N.
* Lines 2..N+1: Line i+1 contains the integer P_i.输出格式：
* Line 1: The minimum total cost of the damage.
Four cows placed in positions: -2, -12, 3, and 7.
The optimal visit order is -2, 3, 7, -12. FJ arrives at position -2 in 2 minutes for a total of 2 dollars in damage for that cow.
He then travels to position 3 (distance: 5) where the cumulative damage is 2 + 5 = 7 dollars for that cow.
He spends 4 more minutes to get to 7 at a cost of 7 + 4 = 11 dollars for that cow.
Finally, he spends 19 minutes to go to -12 with a cost of 11 + 19 = 30 dollars.
The total damage is 2 + 7 + 11 + 30 = 50 dollars.