Exercise 3 solution

    import heapq
    import itertools

    class Scheduler:
        def __init__(self):
            self._heap = []
            self._counter = itertools.count()
            self._cancelled = set()

        def add(self, priority, name):
            n = next(self._counter)
            heapq.heappush(self._heap, (priority, n, name))
            return n                       # a ticket for cancel()

        def cancel(self, ticket):
            self._cancelled.add(ticket)

        def next_task(self):
            while self._heap:
                priority, n, name = heapq.heappop(self._heap)
                if n in self._cancelled:
                    self._cancelled.discard(n)
                    continue
                return name
            return None

    s = Scheduler()
    s.add(2, "write report")
    s.add(1, "fix outage")
    t = s.add(3, "tidy desk")
    s.add(1, "restart server")
    s.add(2, "send invoice")
    s.cancel(t)
    while True:
        task = s.next_task()
        if task is None:
            break
        print(task)

Output:
    fix outage
    restart server
    write report
    send invoice

Lower numbers run first. The counter in the middle of each tuple has two
jobs: tasks with equal priority come out in the order they were added
("fix outage" before "restart server"), and Python never has to compare the
names. Removing an item from the middle of a heap is awkward, so cancelling
is done lazily: the ticket is remembered and the stale entry is simply
skipped when it eventually reaches the top. "tidy desk" never appears.