Convert heapq.py to a C implementation.

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diff --git a/Lib/heapq.py b/Lib/heapq.py
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-# -*- coding: Latin-1 -*-
-
-"""Heap queue algorithm (a.k.a. priority queue).
-
-Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
-all k, counting elements from 0.  For the sake of comparison,
-non-existing elements are considered to be infinite.  The interesting
-property of a heap is that a[0] is always its smallest element.
-
-Usage:
-
-heap = []            # creates an empty heap
-heappush(heap, item) # pushes a new item on the heap
-item = heappop(heap) # pops the smallest item from the heap
-item = heap[0]       # smallest item on the heap without popping it
-heapify(x)           # transforms list into a heap, in-place, in linear time
-item = heapreplace(heap, item) # pops and returns smallest item, and adds
-                               # new item; the heap size is unchanged
-
-Our API differs from textbook heap algorithms as follows:
-
-- We use 0-based indexing.  This makes the relationship between the
-  index for a node and the indexes for its children slightly less
-  obvious, but is more suitable since Python uses 0-based indexing.
-
-- Our heappop() method returns the smallest item, not the largest.
-
-These two make it possible to view the heap as a regular Python list
-without surprises: heap[0] is the smallest item, and heap.sort()
-maintains the heap invariant!
-"""
-
-# Original code by Kevin O'Connor, augmented by Tim Peters
-
-__about__ = """Heap queues
-
-[explanation by François Pinard]
-
-Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for
-all k, counting elements from 0.  For the sake of comparison,
-non-existing elements are considered to be infinite.  The interesting
-property of a heap is that a[0] is always its smallest element.
-
-The strange invariant above is meant to be an efficient memory
-representation for a tournament.  The numbers below are `k', not a[k]:
-
-                                   0
-
-                  1                                 2
-
-          3               4                5               6
-
-      7       8       9       10      11      12      13      14
-
-    15 16   17 18   19 20   21 22   23 24   25 26   27 28   29 30
-
-
-In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'.  In
-an usual binary tournament we see in sports, each cell is the winner
-over the two cells it tops, and we can trace the winner down the tree
-to see all opponents s/he had.  However, in many computer applications
-of such tournaments, we do not need to trace the history of a winner.
-To be more memory efficient, when a winner is promoted, we try to
-replace it by something else at a lower level, and the rule becomes
-that a cell and the two cells it tops contain three different items,
-but the top cell "wins" over the two topped cells.
-
-If this heap invariant is protected at all time, index 0 is clearly
-the overall winner.  The simplest algorithmic way to remove it and
-find the "next" winner is to move some loser (let's say cell 30 in the
-diagram above) into the 0 position, and then percolate this new 0 down
-the tree, exchanging values, until the invariant is re-established.
-This is clearly logarithmic on the total number of items in the tree.
-By iterating over all items, you get an O(n ln n) sort.
-
-A nice feature of this sort is that you can efficiently insert new
-items while the sort is going on, provided that the inserted items are
-not "better" than the last 0'th element you extracted.  This is
-especially useful in simulation contexts, where the tree holds all
-incoming events, and the "win" condition means the smallest scheduled
-time.  When an event schedule other events for execution, they are
-scheduled into the future, so they can easily go into the heap.  So, a
-heap is a good structure for implementing schedulers (this is what I
-used for my MIDI sequencer :-).
-
-Various structures for implementing schedulers have been extensively
-studied, and heaps are good for this, as they are reasonably speedy,
-the speed is almost constant, and the worst case is not much different
-than the average case.  However, there are other representations which
-are more efficient overall, yet the worst cases might be terrible.
-
-Heaps are also very useful in big disk sorts.  You most probably all
-know that a big sort implies producing "runs" (which are pre-sorted
-sequences, which size is usually related to the amount of CPU memory),
-followed by a merging passes for these runs, which merging is often
-very cleverly organised[1].  It is very important that the initial
-sort produces the longest runs possible.  Tournaments are a good way
-to that.  If, using all the memory available to hold a tournament, you
-replace and percolate items that happen to fit the current run, you'll
-produce runs which are twice the size of the memory for random input,
-and much better for input fuzzily ordered.
-
-Moreover, if you output the 0'th item on disk and get an input which
-may not fit in the current tournament (because the value "wins" over
-the last output value), it cannot fit in the heap, so the size of the
-heap decreases.  The freed memory could be cleverly reused immediately
-for progressively building a second heap, which grows at exactly the
-same rate the first heap is melting.  When the first heap completely
-vanishes, you switch heaps and start a new run.  Clever and quite
-effective!
-
-In a word, heaps are useful memory structures to know.  I use them in
-a few applications, and I think it is good to keep a `heap' module
-around. :-)
-
---------------------
-[1] The disk balancing algorithms which are current, nowadays, are
-more annoying than clever, and this is a consequence of the seeking
-capabilities of the disks.  On devices which cannot seek, like big
-tape drives, the story was quite different, and one had to be very
-clever to ensure (far in advance) that each tape movement will be the
-most effective possible (that is, will best participate at
-"progressing" the merge).  Some tapes were even able to read
-backwards, and this was also used to avoid the rewinding time.
-Believe me, real good tape sorts were quite spectacular to watch!
-From all times, sorting has always been a Great Art! :-)
-"""
-
-__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace']
-
-def heappush(heap, item):
-    """Push item onto heap, maintaining the heap invariant."""
-    heap.append(item)
-    _siftdown(heap, 0, len(heap)-1)
-
-def heappop(heap):
-    """Pop the smallest item off the heap, maintaining the heap invariant."""
-    lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
-    if heap:
-        returnitem = heap[0]
-        heap[0] = lastelt
-        _siftup(heap, 0)
-    else:
-        returnitem = lastelt
-    return returnitem
-
-def heapreplace(heap, item):
-    """Pop and return the current smallest value, and add the new item.
-
-    This is more efficient than heappop() followed by heappush(), and can be
-    more appropriate when using a fixed-size heap.  Note that the value
-    returned may be larger than item!  That constrains reasonable uses of
-    this routine.
-    """
-    returnitem = heap[0]    # raises appropriate IndexError if heap is empty
-    heap[0] = item
-    _siftup(heap, 0)
-    return returnitem
-
-def heapify(x):
-    """Transform list into a heap, in-place, in O(len(heap)) time."""
-    n = len(x)
-    # Transform bottom-up.  The largest index there's any point to looking at
-    # is the largest with a child index in-range, so must have 2*i + 1 < n,
-    # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
-    # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
-    # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
-    for i in reversed(xrange(n//2)):
-        _siftup(x, i)
-
-# 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
-# is the index of a leaf with a possibly out-of-order value.  Restore the
-# heap invariant.
-def _siftdown(heap, startpos, pos):
-    newitem = heap[pos]
-    # Follow the path to the root, moving parents down until finding a place
-    # newitem fits.
-    while pos > startpos:
-        parentpos = (pos - 1) >> 1
-        parent = heap[parentpos]
-        if parent <= newitem:
-            break
-        heap[pos] = parent
-        pos = parentpos
-    heap[pos] = newitem
-
-# The child indices of heap index pos are already heaps, and we want to make
-# a heap at index pos too.  We do this by bubbling the smaller child of
-# pos up (and so on with that child's children, etc) until hitting a leaf,
-# then using _siftdown to move the oddball originally at index pos into place.
-#
-# We *could* break out of the loop as soon as we find a pos where newitem <=
-# both its children, but turns out that's not a good idea, and despite that
-# many books write the algorithm that way.  During a heap pop, the last array
-# element is sifted in, and that tends to be large, so that comparing it
-# against values starting from the root usually doesn't pay (= usually doesn't
-# get us out of the loop early).  See Knuth, Volume 3, where this is
-# explained and quantified in an exercise.
-#
-# Cutting the # of comparisons is important, since these routines have no
-# way to extract "the priority" from an array element, so that intelligence
-# is likely to be hiding in custom __cmp__ methods, or in array elements
-# storing (priority, record) tuples.  Comparisons are thus potentially
-# expensive.
-#
-# On random arrays of length 1000, making this change cut the number of
-# comparisons made by heapify() a little, and those made by exhaustive
-# heappop() a lot, in accord with theory.  Here are typical results from 3
-# runs (3 just to demonstrate how small the variance is):
-#
-# Compares needed by heapify     Compares needed by 1000 heappops
-# --------------------------     --------------------------------
-# 1837 cut to 1663               14996 cut to 8680
-# 1855 cut to 1659               14966 cut to 8678
-# 1847 cut to 1660               15024 cut to 8703
-#
-# Building the heap by using heappush() 1000 times instead required
-# 2198, 2148, and 2219 compares:  heapify() is more efficient, when
-# you can use it.
-#
-# The total compares needed by list.sort() on the same lists were 8627,
-# 8627, and 8632 (this should be compared to the sum of heapify() and
-# heappop() compares):  list.sort() is (unsurprisingly!) more efficient
-# for sorting.
-
-def _siftup(heap, pos):
-    endpos = len(heap)
-    startpos = pos
-    newitem = heap[pos]
-    # Bubble up the smaller child until hitting a leaf.
-    childpos = 2*pos + 1    # leftmost child position
-    while childpos < endpos:
-        # Set childpos to index of smaller child.
-        rightpos = childpos + 1
-        if rightpos < endpos and heap[rightpos] <= heap[childpos]:
-            childpos = rightpos
-        # Move the smaller child up.
-        heap[pos] = heap[childpos]
-        pos = childpos
-        childpos = 2*pos + 1
-    # The leaf at pos is empty now.  Put newitem there, and bubble it up
-    # to its final resting place (by sifting its parents down).
-    heap[pos] = newitem
-    _siftdown(heap, startpos, pos)
-
-if __name__ == "__main__":
-    # Simple sanity test
-    heap = []
-    data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0]
-    for item in data:
-        heappush(heap, item)
-    sort = []
-    while heap:
-        sort.append(heappop(heap))
-    print sort