possible - sdu
Post on 29-Nov-2021
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idea : try to me asmuch as
possibleout of s :
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A preflowin N - (Vols ,El , A. toil )
is a flow x s - t :
bxlo) E oVo EV - s
( so s is the only outerwith positive
balance )Note that every p
reflow de company
into path and ayah flow ,atoms P
, ,R ,.. Pa
,
Ci,
.
- icp
when each Pi starts in s and
end , in a uwhxv with 8×6140
leinma if x is a pretlow and8×6 ) so thin NK) contain , a @
s 1.pathbxloko
so#in Next
h is a height function w rt x if
h (s ) = n,
htt = o
h ( p ) Eh@It I tpqENG) * I
:c. . in
"
ha - I p'
no such arc
Example of a height function:
distance to f : hurt = distal ,suit )
dist (pit Is distant) H t peg C-
NEI
•-7 a -7 .-7 a→ a→a
pt
posh Cps) precondition beep) so p.tt-
and hip) - h tl I Ei9-
update Xpqexpqtgwhen @ I
g = min 3-8×407 , rpozl
two pooibh outcomes of a push Cpf )
• p becomesbalanced ( poroibhunsstund
pools )
•are pfcfncxlathr @ 1
( pf becamesaturated by the push )
lift@ I prone bxcokoA
- .e- hei
--- F-D -
--
he t - I
- no arc down
④ h @ I ← min } he Itt Ipa ENG )A
nap ÷.
Nod if 8×6 ) sothen v has an out -wish
tour in NH )
(so @ 1 it welldefined )
reason 8×61 so ⇒ @pl -path is NK)
Geuwicpmtlouw-pushalgont.hn#prepwassius-a7V-peVhpk-distnlpit)
(b) listen② XsqE Usf Fsf
C- A
@ I Xijeo forall otherare,
Maintopwhile Io EV -⑤Hs . tbxloko
if Nfl contains an arc rot
with h@ that I then push loot)
eln letter )
④ h remain , a height function during
the algorithm :
• ok initially a, h@iedistnCu.t )
•has only changed when we
left@ I is applied and this
preserve ,the property
• push Co2) may creakthe arc
zero •
→ -5¥.h G) = heel - I
61 X remain , a pretlowpreserved by push operation
@l tf the algorithm terminate,
this X is a maximum flow
. at termination btw ) - o tu EV-Kitt
so X is an⑤ fl - flow
•then in . ⑤fl - path in NH )
n - hsueh.
it---
• HEI - o
such a path would have narcs tf
M FMC th m → x is a mat flow .
-
claim the algorithm does terminate
and it un, atmoot Ocnhm )
operation, .(A) at most 044 lifts in thealgorithmliftCo1increunsh@lbyaf6a.t Iand h@ is Lu - I
must otilhild
Since NCH ha , aI -path
⇒ if initially here Ialways h@ Is Ln - 2
Iv butted at most 2n him,
④ I ocnm) saturating pushes :bound .# time, push (pH is
executed
for a given arc Pf :
^
¥÷ '¥:it:*.
'
before we canpmreduce Xpat9-
next time we push from p tog
hpu has moreand by at least 2 .
at h@Is 2n - I in the whole algorithm
we have at most n pain , fuomp to f
¢ ) O Crim) on sshumhu, pushes :
define OI - ZhouHayao
⇐ OI zoalways
in chilly oIoE2n2contribution , to OI during the algorithm :
• lifts contribute Ocn' ) by @ I
•
saturating pushes contribute
o ( n'm ) by BCI( OCaml satumhn, pushes ,eachcontributing Oln) )
• Each unsatumhn , push decream
Ess at least one !
ConclusionIo E Lu
' and the total
in cream in IT during the
algorithm is Ocam )
so # unsafe mtg pu.hu/iOCu2m)
So 1) the algorithmterminates-
2) us in ,Och'm) operators
pretlow push algorithm un, och'm) operation,
active vertices ( 8×61co )
- all have bro
Given or with 8×6 ) co
-Toa- - - -
find f s -t h@ teh@ Itt
-¥ - and ofENCK)
9-
lift : he ,← mm 3h@HI IKENCH )
-keep adjacency list representationof NEI
info about
root
¥ea÷ .
to
Ahuja 7.7 Fito preflow push ab-
Rule : once v ihr th 8×14 cois
-
Chona,
we keep pushing fromou ht either be , becomes
o
or we litter .
( node examination step )
Do this in Fifo orb in gut
⇐Hlltout
8×40
Partition examination operationsinto phan,
Phani : do node examination for allv that got flow flowsin the initialization
s . !) . bro←8×0
Than i : do node examination
on all nodes inthe hot
yera th phan iy
Any node is processed atmoot
once pr Phan
← /Td topmen in phan ie ,
THE- I
Claim then an at most
2hhen pham inthe algorithm
at oI= max 1 hell 8×610 I
consider the total change of E
during a phan :
Ei - Eit9 Tend of end of
phani phani- I
Cant we performed Z l lift in
phan i .Then I could incream but no
more than 242 over the
whole algorithm
cants no lifts in phani( each or tox from phan
i got
balanced dhnh,the phan )
OI will decrease byat least on
( Every vertexin the list when
phan iend, have heights OI
Conclusion I 2h44 phamT initially pot
hat -- 4
if no @El - path iNCx)
This implies thattotal
number of wnostonihin , push,
isan ' ) :
Each ve thisexamined at
most once per phanand
utmost one on satorn tin ,
push fromu in a phan
So since # pham is Olay
we do 0 ( n' ) uusntvmht
pushes .
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