PAT 1008

PAT1008：电梯调度问题的高效解决方法

最新推荐文章于 2025-09-09 19:58:37 发布

转载最新推荐文章于 2025-09-09 19:58:37 发布 · 45 阅读

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原文链接：http://www.cnblogs.com/boostable/p/pat_1008.html

本文详细介绍了如何通过优化算法解决PAT1008中的电梯调度问题，包括输入数据解析、算法实现及输出结果计算。通过分析电梯运行规律，提出了一种减少总运行时间的策略，旨在提升城市建筑中电梯服务效率。

PAT 1008

1008. Elevator (20)

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:

直接上代码

1 #include <stdio.h>
2
3 int main()
4 {
5      int N;
6      int sum;
7      int i;
8      int start_pos,end_pos;
9      while(scanf( " %d ",&N) != EOF){
10         sum = 0;
11         start_pos = 0;
12          for(i= 0;i<N;++i){
13             scanf( " %d ",&end_pos);
14              if(start_pos < end_pos){
15                 sum = sum + 6 * (end_pos - start_pos);
16             }
17              else{
18                 sum = sum + 4 * (start_pos - end_pos);
19             }
20             start_pos = end_pos;
21         }
22         sum = sum + 5 * N;
23         printf( " %d\n ",sum);
24     }
25      return 0;
26 }

posted on 2014-02-11 18:24 Boostable 阅读( ...) 评论( ...) 编辑收藏

转载于:https://www.cnblogs.com/boostable/p/pat_1008.html