PAT (Advanced Level) 1008 Elevator

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本文详细介绍了一个电梯调度问题的算法实现，电梯在多个楼层间的移动时间计算，包括上行和下行的时间成本，以及在每个楼层停留的时间。通过具体的代码示例，展示了如何计算完成一系列楼层请求所需的总时间。

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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:

AC代码：

#include <iostream>
#include <cmath> 
using namespace std;
int main(){
	int floor[100];
	int N; //电梯所要停下的次数
	int totalTime = 0; //电梯停止的总时间 
	cin >> N;
	for(int i = 0; i < N; i++) {
		cin >> floor[i];
	}
	
	//由于一开始电梯处于0层，所以还要加上0-floor[0]的时间 
	totalTime += 6*floor[0];
	
	for(int j = 0; j < N; j++) {
		if(j == N-1) break; //到达最后一层就不再往下算 
		//判断电梯是上行还是下行，然后再计算时间 
		totalTime += (floor[j+1] -floor[j])>0 ? 6*abs(floor[j+1]-floor[j]) : 4*abs(floor[j+1]-floor[j]);
	} 
	
	//最后还要加上每层停的时间 
	totalTime += 5*N;
	cout << totalTime;
	return 0;
}