PAT 1008 Elevator （20 分）

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本文详细介绍了一个关于电梯调度的算法问题，模拟了城市中最高建筑内仅有的一部电梯如何高效响应楼层请求的过程。通过分析给定的楼层请求列表，计算出完成所有请求所需的总时间，考虑了电梯上下楼的时间成本及在每层停留的时间。文章提供了完整的C++代码实现，帮助读者理解并掌握电梯调度算法的细节。

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1008 Elevator （20 分）

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.
For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.
Input Specification:
Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.
Output Specification:
For each test case, print the total time on a single line.
Sample Input:

3 2 3 1

Sample Output:

解析

模拟题。
可以看看题目给的输入：是怎么算出41的。再模拟这个情况就行了
在这里插入图片描述

Code:

#include<iostream>
#include<vector>
using namespace std;
const unsigned up = 6,down=4,stay=5;
int main()
{
	int N,cost=0,Where=0;
	cin >> N;
	vector<int> floor(N, 0);
	for (int i = 0; i < N; i++)
		cin >> floor[i];
	for (int i = 0; i < N; i++) {
		cost += (floor[i]-Where > 0) ? (floor[i]-Where)*up : abs(floor[i]-Where)*down;
		Where = floor[i];
		cost += stay;
	}
	cout << cost;
}