÷ ... ee

idea : try to me asmuch as

possibleout of s :

Iraj45fill then

xsjeasj

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 .