Mercurial > public > mercurial-scm > hg
view mercurial/ancestor.py @ 7305:c21d236ca897
hgweb: descend empty directories in web view
When a manifest has a series of directories with nothing in them but a single
directory, displaying the entire chain of empty directories allows for
navigation down to the first non-empty directory with a single click.
Because Java links package hierarchy to directory hierarchy, and because Java
conventions include at least three empty directories at the top of this
hierarchy, descending down empty directories is very common in Java web tools.
author | Ry4an Brase <ry4an-hg@ry4an.org> |
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date | Mon, 03 Nov 2008 10:20:28 +0100 |
parents | 532ca442b903 |
children | 8d78fc991b71 |
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# ancestor.py - generic DAG ancestor algorithm for mercurial # # Copyright 2006 Matt Mackall <mpm@selenic.com> # # This software may be used and distributed according to the terms # of the GNU General Public License, incorporated herein by reference. import heapq def ancestor(a, b, pfunc): """ return the least common ancestor of nodes a and b or None if there is no such ancestor. pfunc must return a list of parent vertices """ if a == b: return a # find depth from root of all ancestors visit = [a, b] depth = {} while visit: vertex = visit[-1] pl = pfunc(vertex) if not pl: depth[vertex] = 0 visit.pop() else: for p in pl: if p == a or p == b: # did we find a or b as a parent? return p # we're done if p not in depth: visit.append(p) if visit[-1] == vertex: depth[vertex] = min([depth[p] for p in pl]) - 1 visit.pop() # traverse ancestors in order of decreasing distance from root def ancestors(vertex): h = [(depth[vertex], vertex)] seen = {} while h: d, n = heapq.heappop(h) if n not in seen: seen[n] = 1 yield (d, n) for p in pfunc(n): heapq.heappush(h, (depth[p], p)) def generations(vertex): sg, s = None, {} for g, v in ancestors(vertex): if g != sg: if sg: yield sg, s sg, s = g, {v:1} else: s[v] = 1 yield sg, s x = generations(a) y = generations(b) gx = x.next() gy = y.next() # increment each ancestor list until it is closer to root than # the other, or they match try: while 1: if gx[0] == gy[0]: for v in gx[1]: if v in gy[1]: return v gy = y.next() gx = x.next() elif gx[0] > gy[0]: gy = y.next() else: gx = x.next() except StopIteration: return None