scripts/chg-len.py
author Peter Kovacs <kpeter@inf.elte.hu>
Thu, 12 Nov 2009 23:26:13 +0100
changeset 806 fa6f37d7a25b
parent 422 62c1ed05e83f
permissions -rwxr-xr-x
Entirely rework CapacityScaling (#180)

- Use the new interface similarly to NetworkSimplex.
- Rework the implementation using an efficient internal structure
for handling the residual network. This improvement made the
code much faster (up to 2-5 times faster on large graphs).
- Handle GEQ supply type (LEQ is not supported).
- Handle negative costs for arcs of finite capacity.
(Note that this algorithm cannot handle arcs of negative cost
and infinite upper bound, thus it returns UNBOUNDED if such
an arc exists.)
- Extend the documentation.
     1 #! /usr/bin/env python
     2 #
     3 # This file is a part of LEMON, a generic C++ optimization library.
     4 #
     5 # Copyright (C) 2003-2009
     6 # Egervary Jeno Kombinatorikus Optimalizalasi Kutatocsoport
     7 # (Egervary Research Group on Combinatorial Optimization, EGRES).
     8 #
     9 # Permission to use, modify and distribute this software is granted
    10 # provided that this copyright notice appears in all copies. For
    11 # precise terms see the accompanying LICENSE file.
    12 #
    13 # This software is provided "AS IS" with no warranty of any kind,
    14 # express or implied, and with no claim as to its suitability for any
    15 # purpose.
    16 
    17 import sys
    18 
    19 from mercurial import ui, hg
    20 from mercurial import util
    21 
    22 util.rcpath = lambda : []
    23 
    24 if len(sys.argv)>1 and sys.argv[1] in ["-h","--help"]:
    25     print """
    26 This utility just prints the length of the longest path
    27 in the revision graph from revison 0 to the current one.
    28 """
    29     exit(0)
    30 
    31 u = ui.ui()
    32 r = hg.repository(u, ".")
    33 N = r.changectx(".").rev()
    34 lengths=[0]*(N+1)
    35 for i in range(N+1):
    36     p=r.changectx(i).parents()
    37     if p[0]:
    38         p0=lengths[p[0].rev()]
    39     else:
    40         p0=-1
    41     if len(p)>1 and p[1]:
    42         p1=lengths[p[1].rev()]
    43     else:
    44         p1=-1
    45     lengths[i]=max(p0,p1)+1
    46 print lengths[N]