HDU 1008

最新推荐文章于 2019-03-18 20:48:45 发布

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本文介绍了一个基于电梯调度的算法问题，通过给定一系列电梯请求楼层的列表，计算完成所有请求所需的总时间。考虑到了电梯向上移动、向下移动及停靠时间，提供了完整的C语言实现代码。

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Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50245 Accepted Submission(s): 27678

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

Output
Print the total time on a single line for each test case.

Sample Input
1 2
3 2 3 1
0

Sample Output
17
41

Author
ZHENG, Jianqiang

Source
ZJCPC2004

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注意了：你连续输入相同的楼层也要算时间，即要加5.

#include<stdio.h>
int main()
{
  int n;
  while(scanf("%d",&n)!=EOF&&n&&n<100)
  {int i;
      int target[n];
      for(i=0;i<n;i++)
        {
        scanf("%d",&target[i]);
        if(target[i]>=100) return 0;
        }

      int count=target[0]*6+5;
      for( i=0;i<n-1;i++)
      {
          if(target[i+1]>target[i]) count+=(target[i+1]-target[i])*6+5;
          if(target[i+1]<target[i]) count+=(target[i]-target[i+1])*4+5;
          if(target[i+1]==target[i]) count+=5;
      }
      printf("%d\n",sum);
  }
  return 0;
}