[leetcode] 1599. Maximum Profit of Operating a Centennial Wheel
Description
You are the operator of a Centennial Wheel that has four gondolas, and each gondola has room for up to four people. You have the ability to rotate the gondolas counterclockwise, which costs you runningCost dollars.
You are given an array customers of length n where customers[i] is the number of new customers arriving just before the ith rotation (0-indexed). This means you must rotate the wheel i times before the customers[i] customers arrive. You cannot make customers wait if there is room in the gondola. Each customer pays boardingCost dollars when they board on the gondola closest to the ground and will exit once that gondola reaches the ground again.
You can stop the wheel at any time, including before serving all customers. If you decide to stop serving customers, all subsequent rotations are free in order to get all the customers down safely. Note that if there are currently more than four customers waiting at the wheel, only four will board the gondola, and the rest will wait for the next rotation.
Return the minimum number of rotations you need to perform to maximize your profit. If there is no scenario where the profit is positive, return -1.
Example 1:
Input: customers = [8,3], boardingCost = 5, runningCost = 6
Output: 3
Explanation: The numbers written on the gondolas are the number of people currently there.
1. 8 customers arrive, 4 board and 4 wait for the next gondola, the wheel rotates. Current profit is 4 * $5 - 1 * $6 = $14.
2. 3 customers arrive, the 4 waiting board the wheel and the other 3 wait, the wheel rotates. Current profit is 8 * $5 - 2 * $6 = $28.
3. The final 3 customers board the gondola, the wheel rotates. Current profit is 11 * $5 - 3 * $6 = $37.
The highest profit was $37 after rotating the wheel 3 times.
Example 2:
Input: customers = [10,9,6], boardingCost = 6, runningCost = 4
Output: 7
Explanation:
1. 10 customers arrive, 4 board and 6 wait for the next gondola, the wheel rotates. Current profit is 4 * $6 - 1 * $4 = $20.
2. 9 customers arrive, 4 board and 11 wait (2 originally waiting, 9 newly waiting), the wheel rotates. Current profit is 8 * $6 - 2 * $4 = $40.
3. The final 6 customers arrive, 4 board and 13 wait, the wheel rotates. Current profit is 12 * $6 - 3 * $4 = $60.
4. 4 board and 9 wait, the wheel rotates. Current profit is 16 * $6 - 4 * $4 = $80.
5. 4 board and 5 wait, the wheel rotates. Current profit is 20 * $6 - 5 * $4 = $100.
6. 4 board and 1 waits, the wheel rotates. Current profit is 24 * $6 - 6 * $4 = $120.
7. 1 boards, the wheel rotates. Current profit is 25 * $6 - 7 * $4 = $122.
The highest profit was $122 after rotating the wheel 7 times.
Example 3:
Input: customers = [3,4,0,5,1], boardingCost = 1, runningCost = 92
Output: -1
Explanation:
1. 3 customers arrive, 3 board and 0 wait, the wheel rotates. Current profit is 3 * $1 - 1 * $92 = -$89.
2. 4 customers arrive, 4 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 2 * $92 = -$177.
3. 0 customers arrive, 0 board and 0 wait, the wheel rotates. Current profit is 7 * $1 - 3 * $92 = -$269.
4. 5 customers arrive, 4 board and 1 waits, the wheel rotates. Current profit is 11 * $1 - 4 * $92 = -$357.
5. 1 customer arrives, 2 board and 0 wait, the wheel rotates. Current profit is 13 * $1 - 5 * $92 = -$447.
The profit was never positive, so return -1.
Example 4:
Input: customers = [10,10,6,4,7], boardingCost = 3, runningCost = 8
Output: 9
Explanation:
1. 10 customers arrive, 4 board and 6 wait, the wheel rotates. Current profit is 4 * $3 - 1 * $8 = $4.
2. 10 customers arrive, 4 board and 12 wait, the wheel rotates. Current profit is 8 * $3 - 2 * $8 = $8.
3. 6 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 12 * $3 - 3 * $8 = $12.
4. 4 customers arrive, 4 board and 14 wait, the wheel rotates. Current profit is 16 * $3 - 4 * $8 = $16.
5. 7 customers arrive, 4 board and 17 wait, the wheel rotates. Current profit is 20 * $3 - 5 * $8 = $20.
6. 4 board and 13 wait, the wheel rotates. Current profit is 24 * $3 - 6 * $8 = $24.
7. 4 board and 9 wait, the wheel rotates. Current profit is 28 * $3 - 7 * $8 = $28.
8. 4 board and 5 wait, the wheel rotates. Current profit is 32 * $3 - 8 * $8 = $32.
9. 4 board and 1 waits, the wheel rotates. Current profit is 36 * $3 - 9 * $8 = $36.
10. 1 board and 0 wait, the wheel rotates. Current profit is 37 * $3 - 10 * $8 = $31.
The highest profit was $36 after rotating the wheel 9 times.
Constraints:
- n == customers.length
- 1 <= n <= 105
- 0 <= customers[i] <= 50
- 1 <= boardingCost, runningCost <= 100
分析
题目的意思是：给你一个数组，给定一个customers的数组，boardingCost表示的是停留的代价，runningCost表示的是运行的代价。
最终的利润为：
profit=onboard*boardingCost-times*runningCost
onboard表示当前停留的人数，times表示的是运行的次数。每次停留的人数不超过4人。
如果这个读懂了的话，就好办了，直接把这个过程模拟出来就行了，具体实现过程看代码，我一开始没读懂，看来我的英文水平需要提升了。
代码
class Solution:
def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
onboard=0
curpeople=0
i=0
n=len(customers)
profit=0
times=0
maxprofit=0
while(i<n or curpeople>0):
if(i<n):
curpeople+=customers[i]
i+=1
if(curpeople>4):
onboard+=4
else:
onboard+=curpeople
times+=1
profit=onboard*boardingCost-times*runningCost
if(profit>maxprofit):
maxprofit=profit
maxround=times
if(curpeople>4):
curpeople-=4
else:
curpeople=0
if(maxprofit<=0):
return -1
else:
return maxround