通过将间隔分配给两个不同的处理器，使间隔不重叠

给定间隔列表interval [] ，其中每个间隔包含两个整数L和R ，任务是将间隔分配给两个不同的处理器，以使它们对于每个处理器而言都不是重叠的间隔。要将间隔[i]分配给第一个处理器，请打印“ F”，然后将其分配给第二个处理器，请打印“ S”。

注意：如果没有可能的解决方法，请打印-1。

例子：

Input: interval[] = {{360, 480}, {420, 540}, {600, 660}}
Output: S, F, S
Explanation:
The intervals assigned to processors are –
Intervals of First Processor {{420, 540}}
Intervals of Second Processor {{360, 480}, {600, 660}}
As there are no overlapping intervals for each processor, it will be a valid solution.

Input: interval[] = {{99, 150}, {1, 100}, {100, 301}, {2, 5}, {150, 250}}
Output: S, F, F, S, S
Explanation:
The intervals assigned to processors are –
Intervals of First Processor {{1, 100}, {100, 301}}
Intervals of Second Processor {{99, 150}, {2, 5}, {150, 250}}
As there are no overlapping intervals for each processor, it will be a valid solution.

方法：想法是使用贪婪算法将间隔分配给处理器。
如果处理器的最高结束时间小于或等于某个间隔的开始时间，则可以将此间隔分配给处理器。否则，请检查其他处理器。如果无法将任何时间间隔分配给任何处理器，则没有可能的解决方案。

下面是该方法步骤的说明：

• 与当前问题一样，我们必须根据间隔的顺序进行打印。因此，要保存间隔的顺序，请将间隔与其索引配对。
• 按时间间隔对时间间隔进行排序。即L。
• 遍历时间间隔，并将时间间隔分配给处理器，如下所示：

  if (interval[i][0] >= firstProcessorEndTime)
      answer[interval[i]] = "F"
      firstProcessorEndTime = 
          max(firstProcessorEndTime, interval[i][0])
  else if (interval[i][0] >= secondProcessorEndTime)
      answer[interval[i]] = "S"
      secondProcessorEndTime = 
          max(secondProcessorEndTime, interval[i][0])
  else
      print(-1)
  

下面是上述方法的实现：

Python3

# Python implementation for intervals
# scheduling to two processors such 
# that there are no overlapping intervals
  
# Function to assign the intervals 
# to two different processors
def assignIntervals(interval, n):
      
    # Loop to pair the interval
    # with their indices
    for i in range(n):
        interval[i].append(i)
          
    # sorting the interval by 
    # their startb times
    interval.sort(key = lambda x: x[0])
      
    firstEndTime = -1
    secondEndTime = -1
    fin = ''
    flag = False
      
    # Loop to iterate over the
    # intervals with their start time
    for i in range(n):
        if interval[i][0] >= firstEndTime:
            firstEndTime = interval[i][1]
            interval[i].append('S')
        elif interval[i][0] >= secondEndTime:
            secondEndTime = interval[i][1]
            interval[i].append('F')
        else:
            flag = True
            break
      
    # Condition to check if there
    # is a possible solution
    if flag:
        print(-1)
    else:
        form = ['']*n
        for i in range(n):
            indi = interval[i][2]
            form[indi] = interval[i][3]
        # form = ''.join(form)
        print(form, ", ")
  
# Driver Code    
if __name__ == "__main__":
    intervals = [[360, 480], [420, 540], [600, 660]]
      
    # Function Call
    assignIntervals(intervals, len(intervals))

输出：

['S', 'F', 'S'] ,