PAT-1008 Elevator

最新推荐文章于 2024-12-17 09:09:57 发布

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最新推荐文章于 2024-12-17 09:09:57 发布

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原文链接：http://www.cnblogs.com/chengdalei/p/10720815.html

本文介绍了一个电梯调度问题的算法解决方案，电梯在多个楼层间的移动时间计算，包括上升和下降的时间成本，以及在每一层停留的时间。通过具体示例，展示了如何根据楼层请求列表计算总时间消耗，适用于只有一个电梯的场景。

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:

41

算法说明：

不要存储好楼层再计算，不好控制到底是上升还是下降，会得不全分的。

// 1008 Elevator.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;

int main(int argc, char* argv[])
{
    int floor,stay=0;
    int N;
    int i,time=0;
    
    cin >>N;
    for(i=1;i<=N;i++){
        cin >>floor;
        if(stay==floor){
            time+=5;
            continue;
        }else if(floor>stay){
            time+=(floor-stay)*6+5;
        }else{
            time+=(stay-floor)*4+5;
        }
        stay=floor;
    }
    cout << time <<endl;
    return 0;
}

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转载于:https://www.cnblogs.com/chengdalei/p/10720815.html