杭电1008 Elevator

最新推荐文章于 2020-02-09 09:15:44 发布

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本文介绍了一个基于电梯调度的算法问题，通过分析电梯上下楼层所需时间，计算完成一系列楼层请求所需的总时间。考虑到移动时间和停靠时间，该算法适用于计算特定序列下的电梯运行效率。

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Elevator

Time Limit: 2000/1000 MS(Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31446 Accepted Submission(s): 17014

Problem Description

The highestbuilding in our city has only one elevator. A request list is made up with Npositive numbers. The numbers denote at which floors the elevator will stop, inspecified order. It costs 6 seconds to move the elevator up one floor, and 4seconds to move down one floor. The elevator will stay for 5 seconds at eachstop.

For a given request list, you are to compute the total time spent to fulfillthe requests on the list. The elevator is on the 0th floor at the beginning anddoes not have to return to the ground floor when the requests are fulfilled.

Input

There are multipletest cases. Each case contains a positive integer N, followed by N positivenumbers. All the numbers in the input are less than 100. A test case with N = 0denotes the end of input. This test case is not to be processed.

Output

Print the totaltime on a single line for each test case.

Sample Input

1 2

3 2 3 1

Sample Output

#include <iostream>
using namespace std;
int main()
{
    int n,f[101]={0},i,time=0,total=0;   //n request的个数  f每个请求的层数  i计数 time每次移动的时间
    //total 总时间
    while(cin>>n&&n!=0)                     //输入request 为0时退出
    {
        for(i=0;i<n;i++)
            cin>>f[i+1];                              //输入每个请求
        for(i=0;i<n;i++)
        {
            if(f[i+1]>f[i])
                time=(f[i+1]-f[i])*6;                       //每次移动的时间
            else
                time=(f[i]-f[i+1])*4;
            total+=time;
        }
        total+=5*n;
        cout<<total<<endl;                            //输出总时间
        total=0;time=0;                 //计数器归0
    }
    return 0;
}