algorithm - 活动时间表

Question

我有以下任务。有24小时的时间表。还有一些固定长度的事件。我有一个预定义长度的新事件（例如 1 小时 25 分钟）。任务是要知道 - 我可以在一天中插入多少次这个任务（这个任务不能与其他事件相交。

Answer 1

我们的问题可能被认为有点简单，所以让答案也简单（在 Ruby 中）：

#! /usr/bin/ruby
DAY_START = 0
DAY_END = 24*60
# now, busy periods as array of size-2 arrays:
BUSY_PERIODS = [[9*60, 9*60+30], [18*60+30, 20*60]]
ADDITIONAL_EVENT_LENGTH = 1*60 + 25   # that is, 1 hour and  25 minutes
# in this simple program,
# time will be expressed as a number of minutes of the day
# for more advanced use, try to use "Time" class

# now let define a function to compute the list of free periods
# from a given list of busy periods:
def compute_free_periods( list_of_events )
  # at the begining of the calculation, our whole day is assumed free:
  free_periods = Array[ [ DAY_START, DAY_END] ]
  # now, one by one, let us take away the busy periods:
  list_of_events.each_with_object free_periods do | busy_period, free_periods |
    # we use 'each_with_object' version of 'each' enumerator
    # list of free_periods is passed in as an external object
    # so that we can gradually take each busy period away from it

    # let us first note the end time for the last free period
    # ( [-1] refers to the last element of the list of free periods,
    # subsequent [1] to the second element of the size-2 array )
    last_free_period_end = free_periods[-1][1]

    # and now, let us split this last free period into two by the
    # current busy period (busy periods assumed non-overlapping and
    # sorted in ascending order)

    # first, end time of the last free period is set to the beginning
    # of the busy period that we consider in this iteration:
    free_periods[-1][1] = busy_period[0]

    # then, additional free period is appended to the list of
    # free periods usind << operator, starting when the busy period ends:
    free_periods << [ busy_period[1], last_free_period_end ]
  end
end

# now, let us use the function we just defined:
free_periods = compute_free_periods( BUSY_PERIODS )

# Now, for each free period we will count how many times we can stuff
# in the desired additional event:
free_period_capacities = free_periods.map { |free_period|
  # we are using map function for this, which will return
  # an array of the same length as free_periods, having
  # performed prescribed modifications on each element:

  # first, we calculate how many minutes a free period has
  period_length = free_period[1] - free_period[0]
  # then, we will use whole-number division to get the number
  # of additional events that can fit in:
  period_length / ADDITIONAL_EVENT_LENGTH
  # (in Ruby, whole number division is the default behavior of / operator)
}

# and finally, we sum up the free period capacities for our new event:
total_events_possible = free_period_capacities.reduce( :+ )
# (summing is done elegantly by using reduce function with + operator)

# the last remaining thing we can do to consider program finished is to
puts total_events_possible # print the result on the screen

如果你去掉注释并缩短它，那么程序就会变得很短：

#! /usr/bin/ruby
DAY_START, DAY_END = 0, 24*60
BUSY_PERIODS = [[9*60, 9*60+30], [18*60+30, 20*60]]
ADDITIONAL_EVENT_LENGTH = 1*60 + 25

def compute_free_periods( list_of_events )
  free_periods = Array[ [ DAY_START, DAY_END] ]
  list_of_events.each_with_object free_periods do | busy_period, free_periods |
    last_free_period_end = free_periods[-1][1]
    free_periods[-1][1] = busy_period[0]
    free_periods << [ busy_period[1], last_free_period_end ] end
end

free_periods = compute_free_periods( BUSY_PERIODS )
free_period_capacities = free_periods.map { |free_period|
  period_length = free_period[1] - free_period[0]
  period_length / ADDITIONAL_EVENT_LENGTH }
puts ( total_events_possible = free_period_capacities.reduce( :+ ) )

你看，没有那么多的算法。只是简单的编码。拿起例如。Ole Foul Zed 的教科书Learn Ruby The Hard Way并从练习 0 开始阅读到最后。或者，您可以使用其他语言，例如Python。Ole Zed 具有教学天赋。如果您喜欢这种体验，请不要忘记将其作为电子书购买。