Algorithm 活动时间表

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）。任务是要知道我一天中可以插入此任务多少次（此任务不得与其他事件相交）

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我们的问题可能被认为有点简单化， 因此，让答案也简单化（在Ruby中）：

！/usr/bin/ruby
开始日期=0
结束日期=24*60
#现在，繁忙时段作为大小为2的阵列：
繁忙时段=[[9*60,9*60+30]，[18*60+30,20*60]]
附加事件长度=1*60+25，即1小时25分钟
#在这个简单的程序中,，
#时间将以一天中的分钟数表示
#要获得更高级的使用，请尝试使用“Time”类
#现在让我们定义一个函数来计算空闲时间列表
#从给定的繁忙时段列表中：
def计算空闲时段（事件列表）
#计算开始时，假设我们一整天都是免费的：
空闲时间=数组[[开始日期，结束日期]]
#现在，让我们一个接一个地去掉繁忙的时段：
列出所有事件。每个具有对象空闲时间段的事件都有忙碌时间段、空闲时间段|
#我们使用'each'枚举器的'each_with_object'版本
#空闲时间段列表作为外部对象传入
#这样我们就可以逐渐地把每一段繁忙的时间从它身上移开
#让我们首先注意最后一个空闲时间段的结束时间
#（[-1]表示空闲时间列表的最后一个元素，
#大小为2的数组的第二个元素的后续[1]）
最后一个空闲时段结束=空闲时段[-1][1]
#现在，让我们把最后一个自由期按
#当前繁忙时段（假定繁忙时段不重叠且
#按升序排序）
#首先，将最后一个空闲时间段的结束时间设置为开始时间
在这个迭代中我们考虑的忙碌周期：
空闲时间[-1][1]=繁忙时间[0]
#然后，附加的空闲时间将附加到
“自由周期”和“你尝试了什么？我尝试了微不足道的解决方案，但现在我寻找特殊的ALGOS。没有任何算法可以考虑，因为已经天真的算法完成了O（n）：））中的任务。但是，等等，可能有一件事-牺牲内存以提高性能，将计数作为每个空闲时间段的一个数组来维护，当您的日常日程发生变化时，只更新那些空闲时间段，而这些时间段已经发生了变化。不过，除非你的一天中有像DNA这样的百万个片段，否则在那里收获不多
#! /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

#! /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( :+ ) )