possible  SDU
Transcript of possible  SDU
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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 38×407 , 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
 FD 

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)
Geuwicpmtlouwpushalgont.hn#prepwassiusa7VpeVhpkdistnlpit)
(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, [email protected] )
•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 EVKitt
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 [email protected] 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 .