--- /dev/null
+\input lecnotes.tex
+
+\prednaska{3}{Dinicùv algoritmus}{(zapsala Markéta Popelová)}
+
+
+Na~minulé pøedná¹ce jsme si~ukázali Fordùv-Fulkersonùv algoritmus. Tento algoritmus hledal maximální tok tak, ¾e zaèal s~tokem nulovým a~postupnì ho zvìt¹oval. Pro~ka¾dé zvìt¹ení potøeboval v~síti najít cestu, na~které mají v¹echny hrany kladnou rezervu (po takovéto cestì mù¾eme poslat více, ne¾ po~ní aktuálnì teèe). Ukázali jsme, ¾e pokud takováto cesta existuje, jde tok vylep¹it (zvìt¹it). Zároveò pokud tok jde vylep¹it, pak takováto cesta existuje. Dokázali jsme, ¾e pro~racionální kapacity je algoritmus koneèný a~najde maximální tok.
+
+Fordùv-Fulkersonùv algoritmus má ov¹em znaèné nevýhody. Funguje pouze pro~racionální kapacity a~je pomìrnì pomalý. Nyní si~uká¾eme jiný algoritmus, který nevylep¹uje tok pomocí cest, ale pomocí tokù\dots Budeme k~tomu potøebovat sí» rezerv.
+
+\s{Definice:} {\I Sí» rezerv} k~toku~$f$ v~síti $S=(V,E,z,s,c)$ je sí» $R=(S,f)=(V,E,z,s,r)$, kde~$r(e)$ je rezerva hrany~$e$ v~toku~$f$.
+
+\s{Konvence:} Pro~hranu~$e$ znaèí~$\overleftarrow{e}$ hranu opaènou. Napø. pokud $e=uv$, tak $\overleftarrow{e}=vu$.
+
+Je dùle¾ité si~uvìdomit, ¾e sí» rezerv je závislá jak na~pùvodní síti~$S$, tak na~nìjakém toku~$f$ v~síti~$S$. Sí» rezerv~$R$ se~pak od sítì~$S$ li¹í pouze kapacitami -- sí»~$R$ má jako kapacitu hrany rezervu hrany v pùvodní síti. Pro~pøipomenutí: rezervu hrany~$e$ v~síti $S=(V,E,z,s,c)$ s~tokem~$f$ jsme si~definovali jako $r(e)=c(e) - f(e) + f(\overleftarrow{e})$.
+
+Ne¾ si~uká¾eme samotný algoritmus, doká¾eme si~následující lemma.
+
+\s{Lemma:} Pro~ka¾dý tok~$f$ v~síti~$S$ a~pro~ka¾dý tok~$g$ v~síti $R=(S,f)$ lze v~èase $\O(m)$ nalézt tok~$f'$ v~síti~$S$ takový, ¾e $\vert f' \vert = \vert f \vert + \vert g \vert$.
+
+\proof
+
+Dùkaz rozdìlíme do~tøí krokù. V~prvním kroku si~uká¾eme, jak budeme tok~$f'$ v~síti~$S$ konstruovat. V~druhém kroku doká¾eme, ¾e takto zkonstruované~$f'$ je opravdu tok. A~nakonec uká¾eme, ¾e splòuje po¾adovanou vlastnost, tedy ¾e jeho velikost je souèet velikostí tokù~$f$ a~$g$.
+
+\>{\it 1. konstrukce~$f'$}
+
+\noindent
+Pro ka¾dou dvojici hran~$e, \overleftarrow{e}$ urèíme~$f'(e)$ a~$f'(\overleftarrow{e})$ následovnì:
+
+\itemize\ibull
+\:Pokud~$g(e) = g(\overleftarrow{e}) = 0$, pak nastavme:
+ \itemize\ibull
+ \:$f'(e) := f(e)$,
+ \:$f'(\overleftarrow{e}) := f(\overleftarrow{e})$.
+ \endlist
+
+\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) = 0$, pak polo¾me:
+ \itemize\ibull
+ \:$\varepsilon := \min (g(e), f(\overleftarrow{e}))$,
+ \:$f'(e) := f(e) + g(e) - \varepsilon$,
+ \:$f'(\overleftarrow{e}) := f(\overleftarrow{e}) - \varepsilon$.
+ \endlist
+
+\:Pøípad~$g(e) = 0$ a~$g(\overleftarrow{e}) > 0$ vyøe¹íme obdobnì.
+
+\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) > 0$, pak odeèteme od toku~$g$ cirkulaci po cyklu tvoøeném hranami~$e$ a $\overleftarrow{e}$:
+ \itemize\ibull
+ \:$\delta := \min (g(e),g(\overleftarrow{e}))$,
+ \:$g'(e) := g(e) - \delta$,
+ \:$g'(\overleftarrow{e}) := g(\overleftarrow{e}) - \delta$.
+ \endlist
+
+ Tok $g'$ nyní spadá pod nìkterý z~pøedchozích pøípadù, které u¾ umíme vyøe¹it.
+
+\endlist
+
+\>{\it 2. $f'$ je tok}
+
+\numlist{\ndotted}
+
+\:Nejdøíve ovìøme první podmínku: $\forall e \in E: 0 \leq f(e) \leq c(e)$. Vezmìme libovolnou hranu~$e \in E$. Podle toho, co teèe po~hranách~$e$ a~$\overleftarrow{e}$ v~toku~$g$, jsme rozdìlili konstrukci toku na~tøi pøípady:
+
+ \numlist{\ndotted}
+
+ \:Pokud po~hranách~$e$ a~$\overleftarrow{e}$ netekl ¾ádný tok~$g$, pak jsme nastavili $f'(e) := f(e)$ a~$f'(\overleftarrow{e}) := f(\overleftarrow{e})$. Tedy pokud~$f$ dodr¾oval kapacity, tak pro~$f'$ musí platit to samé.
+
+
+ \:Pokud po~hranì~$e$ tekl tok~$g$ nenulový a~po opaèné nulový, tak jsme zvolili: $f'(e) := f(e) + g(e) - \varepsilon$. Víme, ¾e jsme si~$\varepsilon$ vybrali tak, ¾e $\varepsilon \leq g(e)$. Proto $f'(e) \geq 0$.
+
+ Teï ovìøme, ¾e $f'(e) \leq c(e)$. V~pøípadì, ¾e $\varepsilon = g(e)$, tak $f'(e) = f(e) \leq c(e)$. V~opaèném pøípadì platí, ¾e $\varepsilon = f(\overleftarrow{e})$. Pak ov¹em
+ $$f'(e) = f(e) + g(e) - f(\overleftarrow{e}) \leq $$
+ $$\leq f(e) + \left[ c(e) - f(e) + f(\overleftarrow{e}) \right] - f(\overleftarrow{e}) = c(e).$$
+ Vyu¾ili jsme, ¾e~$g$ je tok v~síti rezerv, tedy $g(e) \leq c(e) - f(e) + f(\overleftarrow{e})$.
+
+ Pro tok $f'(\overleftarrow{e})$ platí, ¾e $\varepsilon \leq f(\overleftarrow{e})$. Proto $f'(\overleftarrow{e}) = f(\overleftarrow{e}) - \varepsilon \geq 0$. Zároveò $f'(\overleftarrow{e}) \leq f(\overleftarrow{e}) \leq c(\overleftarrow{e})$.
+
+ Tím jsme dokázali, ¾e~$f'(e)$ i~$f'(\overleftarrow{e})$ dodr¾ují kapacity.
+
+ \:V posledním pøípadì tekl po~obou hranách kladný tok~$g$. Men¹í tok z~$g(e)$ a~$g(\overleftarrow{e})$ jsme vynulovali a~od vìt¹ího odeèetli ten men¹í. Tok~$g'(e)$ a~$g'(\overleftarrow{e})$ tedy zùstal korektní a~tok~$f'$ u¾ konstruujeme podle pøedchozího pøípadu.
+
+ \endlist
+
+\:Teï musíme je¹tì dokázat, ¾e nový tok neporu¹uje Kirchhoffovy zákony: $$\forall v~\in V \setminus \{z,s\}: f'^\Delta(v)=0.$$
+ % neboli $$\forall v~\in V \setminus \{z,s\}: \sum_{u: uv \in E}{f'(uv)}=\sum_{u: vu \in E}{f'(vu)}.$$
+
+ Vezmìme si~libovolnou hranu~$e = uv \in E$. Uvìdomme si, ¾e pøi~pøechodu z~$f(e)$ na~$f'(e)$ a~z~$f(\overleftarrow{e})$ na~$f'(\overleftarrow{e})$ bylo:
+ \itemize\idot
+ \:$f^\Delta(u)$ sní¾eno o~$g(e)$
+ \:$f^\Delta(v)$ zvý¹eno o~$g(e)$.
+ \endlist
+ Seèteme-li úpravy na v¹ech hranách, dostaneme: $$f'^\Delta(v) = f^\Delta(v) + \sum_{u:uv \in E} g(uv) - \sum_{u:vu \in E} g(vu) =$$ $$= f^\Delta(v) + g^+(v) - g^-(v) = f^\Delta(v) + g^\Delta(v).$$
+
+ Jeliko¾~$f$ byl tok, tak $f^\Delta(v) = 0$ a jeliko¾~$g$ byl tok, tak $g^\Delta(v) = 0$. Proto $f'^\Delta(v) = f^\Delta(v) + g^\Delta(v) = 0$.
+
+\endlist
+
+Tím jsme dokázali, ¾e~$f'$ je tok v~síti~$S$.
+
+\>{\it 3. $\vert f' \vert = \vert f \vert + \vert g \vert$}
+
+Pou¾ijme vztah pro souèet pøebytkù z pøedchozího kroku:
+$$\vert f' \vert = f'^\Delta(s) = f^\Delta(s) + g^\Delta(s) = \vert f \vert + \vert g \vert.$$
+\qed
+
+
+Pro algoritmus budeme potøebovat vybírat kvalitní toky~$g$ v~síti rezerv. Pokud se~nám to bude daøit, bude se~tok~$f'$ rychle zvìt¹ovat, a¾ bychom mohli dojít k~maximálnímu toku. Nejlépe by se~nám hodily co nejvìt¹í toky v~síti rezerv. Kdybychom si~dali za cíl najít v¾dy maximální tok v~síti rezerv, výsledek by byl sice krásný (dostali bychom tak rovnou i~maximální tok v~pùvodní síti), ale problém hledání maximálního toku bychom pouze pøenesli na~jinou sí». Na¹e po¾adavky na~tento tok budou tedy takové, aby byl dostateènì velký, ale abychom bìhem jeho hledání nestrávili moc èasu. Podívejme se, jak se~s~tímto problémem vyrovná {\I Dinicùv algoritmus}. Nejdøíve si~ale zadefinujme nìkolik pojmù.
+
+\s{Definice:} Tok~$f$ je {\I blokující}, jestli¾e pro~ka¾dou orientovanou cestu~$P$ ze~$z$ do~$s$ existuje hrana~$e \in P$ taková, ¾e $f(e) = c(e)$.
+
+\s{Definice:} Sí» je {\I vrstevnatá (proèi¹tìná)}, kdy¾ v¹echny vrcholy a~hrany le¾í na~nejkrat¹ích cestách ze~$z$ do~$s$.
+
+Dinicùv algoritmus zaèíná s~nulovým tokem. Potom v¾dy podle toku~$f$ sestrojí sí» rezerv a~v~ní vyma¾e hrany s~nulovou rezervou. Pokud v~této promazané síti rezerv neexistuje cesta ze~zdroje do~stoku, tak skonèí a~prohlásí tok~$f$ za maximální. Jinak proèistí sí» rezerv tak, aby se~z ní stala vrstevnatá sí» (rozdìlí vrcholy do~vrstev podle vzdálenosti od zdroje a~odstraní pøebyteèné hrany). Ve~vrstevnaté síti najde blokující tok, pomocí nìho¾ zlep¹í tok~$f$. Pak opìt pokraèuje sestrojením sítì rezerv na~tomto vylep¹eném toku~$f$ atd.
+
+\figure{dinic-cistasit.eps}{Proèi¹tìná sí» rozdìlená do~vrstev}{0.4\hsize}
+
+\s{Algoritmus (Dinicùv)}
+
+\algo
+\:$f \leftarrow$ nulový tok.
+\:Sestrojíme sí» rezerv~$R$ a~sma¾eme $e: r(e) = 0$.
+\:$l \leftarrow$ délka nejkrat¹í cesty ze~$z$ do~$s$ v~$R$.
+\:Pokud $l = \infty$, zastavíme se~a vrátíme~$f$.
+\:Proèistíme sí» $R \rightarrow$ sí»~$C$.
+\:$g \leftarrow$ blokující tok v~$C$.
+\:Zlep¹íme tok~$f$ pomocí~$g$.
+\:GOTO 2.
+\endalgo
+
+\s{Pozorování:} Pokud se~algoritmus zastaví, vydá maximální tok.
+
+\proof
+Víme, ¾e~$f$ je stále korektní tok (jediné, jak ho mìníme je pøièítání toku~$g$, co¾ je, jak jsme si~dokázali, \uv{ne¹kodná operace}). Jakmile neexistuje cesta ze~$z$ do~$s$ v~$R$, tak je $f$ maximální tok, nebo» v~tuto dobu by se~zastavil (a vydal maximální tok) i~Fordùv-Fulkersonùv algoritmus, který je korektní.
+\qed
+
+A teï je¹tì musíme ujasnit, jak budeme èistit sí» rezerv a~vybírat blokující tok~$g$.
+
+\s{Algoritmus proèi¹tìní sítì rezerv}
+
+\algo
+\:Rozdìlíme vrcholy do~vrstev podle vzdálenosti od~$z$.
+\:Odstraníme vrstvy za~$s$ (tedy vrcholy, které jsou od~$z$ vzdálenìj¹í ne¾~$s$), hrany do~minulých vrstev a~hrany uvnitø vrstev.
+\:Odstraníme \uv{slepé ulièky}, tedy vrcholy s~$\deg^{out}(v) = 0$, a~to opakovanì pomocí fronty. (Nejdøíve zaøadíme do~fronty v¹echny vrcholy s~ $\deg^{out}(v) = 0$. Pak dokud není fronta prázdná, v¾dy vybereme vrchol~$v$ z~fronty, odstraníme~$v$ a~v¹echny hrany~$uv$. Pro~ka¾dý takový vrchol~$u$ zkontrolujeme, zda se~tím nesní¾il výstupní stupeò vrcholu~$u$ na~nulu ($\deg^{out}(u) = 0$). Pokud sní¾il, tak ho zaøadíme do~fronty.)
+\endalgo
+
+\figure{dinic-neprocistenasit.eps}{Neproèi¹tìná sí». Obsahuje zpìtné hrany, hrany uvnitø vrstvy a~slepé ulièky.}{0.45\hsize}
+
+Hledání blokujícího toku zaèneme s~tokem nulovým. Pak vezmeme v¾dy orientovanou cestu ze~zdroje do~stoku v~síti~$C$. V~této cestì najdeme hranu s~nejni¾¹í hodnotou výrazu $r(e) - g(e)$ (neboli $c(e) - f(e)$ v~pùvodní síti). Tuto hodnotu oznaèíme~$\varepsilon$. Pak ke~ v¹em hranám na~této cestì pøièteme~$\varepsilon$. Pokud tok~$g$ na~nìjaké hranì dosáhne kapacity hrany, co¾ je zde~$r(e)$, tak hranu vyma¾eme. Následnì sí» doèistíme, aby splòovala podmínky vrstevnaté sítì. A~pokud je¹tì existuje nìjaká orientovaná cesta ze~zdroje do~stoku, tak opìt pokraèujeme s~touto cestou.
+
+\s{Algoritmus hledání blokujícího toku}
+
+\algo
+\:$g \leftarrow$ nulový tok.
+\:Dokud existuje orientovaná cesta~$P$ ze~$z$ do~$s$ v~$C$, opakuj:
+\::$\varepsilon \leftarrow \min_{e \in P} (r(e) - g(e))$.
+\::Pro~$\forall e \in P: g(e) \leftarrow g(e) + \varepsilon$.
+\:::Pokud $g(e) = r(e)$, sma¾eme~$e$ z~$C$.
+\::Doèistíme sí» zase pomocí fronty.
+\endalgo
+
+\s{Èasová slo¾itost} Rozeberme si~jednotlivé kroky algoritmu.
+
+\numlist{\ndotted}
+\:Inicializace toku~$f$ \dots $\O(m)$.
+\:Sestrojení sítì rezerv a~smazání hran s~nulovou rezervou \dots $\O(m + n)$.
+\:Najití nejkrat¹í cesty (prohledáváním do~¹íøky) \dots $\O(m + n)$.
+\:Zkontrolování délky nejkrat¹í cesty \dots $\O(1)$.
+\:Proèi¹tìní sítì \dots $\O(m + n)$.
+ \numlist{\ndotted}
+ \:Rozdìlení vrcholù do~vrstev -- provedlo ji¾ prohledávání do~¹íøky \dots $\O(1)$.
+ \:Odstranìní nìkterých hran \dots $\O(m + n)$.
+ \:Odstranìní \uv{slepých ulièek} pomocí fronty -- ka¾dou hranu odstraníme nejvý¹e jedenkrát, ka¾dý vrchol se~dostane do~fronty nejvý¹e jedenkrát \dots $\O(m + n)$.
+ \endlist
+\:Najití blokujícího toku~$g$ \dots $\O(m \cdot n)$.
+ \numlist{\ndotted}
+ \:Inicializace toku~$g$ \dots $\O(m)$.
+ \:Najití orientované cesty v~proèi¹tìné síti rezerv (staèí vzít libovolnou cestu ze~zdroje, nebo» ka¾dá z~nich v~této síti vede do~stoku) \dots $\O(n)$.
+ \:Výbìr minima z~výrazu $r(e) - g(e)$ pøes v¹echny hrany cesty -- ta mù¾e být dlouhá nejvý¹e~$n$ \dots $\O(n)$.
+ \:Pøepoèítání v¹ech hran cesty \dots $\O(n)$.
+ \:Smazání hran cesty, jejich¾ tok~$g(e)$ se~zvý¹il na~hodnotu~$r(e)$ \dots $\O(n)$.
+ \:Doèi¹»ování vyøe¹me zvlá¹».
+ \endlist
+
+ Vnitøní cyklus (kroky 2 a¾ 6) provedeme nejvý¹e~$m$ krát, nebo» pøi~ka¾dém prùchodu vyma¾eme alespoò jednu hranu (tak jsme si~volili~$\varepsilon$).
+
+ Èi¹tìní bìhem celého hledání blokujícího toku~$g$ v~proèi¹tìné síti rezerv trvá dohromady $\O(m + n)$, nebo» ka¾dou hranu a~vrchol sma¾eme nejvý¹e jedenkrát.
+
+ Najití blokujícího toku bude tedy trvat $\O(m \cdot n + (m + n)) = \O(m \cdot n)$.
+
+\:Zlep¹ení toku~$f$ pomocí toku~$g$ \dots $\O(m)$.
+\:Skok na~2. krok \dots $\O(1)$.
+\endlist
+
+Zbývá nám jen urèit, kolikrát projdeme vnìj¹ím cyklem (fází). Doká¾eme si~lemma, ¾e hodnota~$l$ vzroste mezi prùchody vnìj¹ím cyklem (fázemi) alespoò o~1. Z~toho plyne, ¾e vnìj¹ím cyklem mù¾eme projít nejvý¹e $n$-krát, nebo» cesta v síti na~$n$ vrcholech mù¾e být dlouhá nejvý¹e $n$.
+
+Uvìdomme si, ¾e uvnitø vnìj¹ího cyklu pøevládá èlen $\O(m \cdot n)$, tak¾e celková èasová slo¾itost bude $\O(n^2 \cdot m)$.
+
+\s{Lemma:} Hodnota~$l$ (délka nejkrat¹í cesty ze~$z$ do~$s$ v~proèi¹tìné síti) vzroste mezi fázemi alespoò o~1.
+
+\proof
+
+Podíváme se~na~prùbìh jednoho prùchodu vnìj¹ím cyklem. Délku aktuálnì nejkrat¹í cesty ze~zdroje do~stoku oznaème~$l$. V¹echny pùvodní cesty délky~$l$ se~bìhem prùchodu zaruèenì nasytí, proto¾e tok~$g$ je blokující. Musíme v¹ak dokázat, ¾e nemohou vzniknout ¾ádné nové cesty délky~$l$ nebo men¹í. V~síti rezerv toti¾ mohou hrany nejen
+ubývat, ale i~pøibývat: pokud po¹leme tok po~hranì, po~které je¹tì nic neteklo, tak v~protismìru z~dosud nulové rezervy vyrobíme nenulovou. Rozmysleme si~tedy, jaké hrany mohou pøibývat.
+
+Hrany mohou pøibývat jen tehdy, kdy¾ jsme po~opaèné hranì nìco poslali. Ale my nìco posíláme po~hranách pouze z~vrstvy do~té následující. Hrany tedy pøibývají do~minulé vrstvy.
+
+Vznikem nových hran by proto mohly vzniknout nové cesty ze~zdroje do~stoku, které pou¾ívají zpìtné hrany. Jen¾e cesta ze~zdroje do~stoku, která pou¾ije zpìtnou hranu, musí alespoò jednou skoèit o~vrstvu zpìt a~nikdy nemù¾e skoèit o~více ne¾ jednu vrstvu dopøedu, a~proto je její délka alespoò $l+2$. Pokud cesta novou zpìtnou hranu nepou¾ije, má buï délku~$> l$, co¾ je v~poøádku, nebo má délku~$= l$, pak je zablokovaná.
+
+Tím je lemma dokázáno.
+\qed
+
+\figure{dinic-cestashranouzpet.eps}{Cesta u¾ívající novou zpìtnou hranu}{0.4\hsize}
+
+V¹echna dokázaná tvrzení mù¾eme shrnout do~následující vìty:
+
+\s{Vìta:} Dinicùv algoritmus najde maximální tok v~èase $\O(n^2\cdot m)$.
+
+\s{Poznámka:} Algoritmus se~chová hezky na~sítích s~malými celoèíselnými kapacitami, ale kupodivu i~na~rùzných jiných sítích. Èasto se~pou¾ívá, nebo» se~chová efektivnì. A~je mnoho zpùsobù, jak ho je¹tì vylep¹ovat, èi odhadovat ni¾¹í slo¾itost na~speciálních sítích.
+
+\s{Poznámka:} Algoritmus nevy¾aduje racionální kapacity! Dal¹í z~dùvodù, proè maximální tok existuje i~v~síti s~iracionálními kapacitami.
+
+
+%k,s,v,na,do,ke,pro,pøi,a,u,i,po,
+
+\bye
--- /dev/null
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-\input lecnotes.tex
-
-\prednaska{2}{Dinicùv algoritmus}{(zapsala Markéta Popelová)}
-
-
-Na~minulé pøedná¹ce jsme si~ukázali Fordùv-Fulkersonùv algoritmus. Tento algoritmus hledal maximální tok tak, ¾e zaèal s~tokem nulovým a~postupnì ho zvìt¹oval. Pro~ka¾dé zvìt¹ení potøeboval v~síti najít cestu, na~které mají v¹echny hrany kladnou rezervu (po takovéto cestì mù¾eme poslat více, ne¾ po~ní aktuálnì teèe). Ukázali jsme, ¾e pokud takováto cesta existuje, jde tok vylep¹it (zvìt¹it). Zároveò pokud tok jde vylep¹it, pak takováto cesta existuje. Dokázali jsme, ¾e pro~racionální kapacity je algoritmus koneèný a~najde maximální tok.
-
-Fordùv-Fulkersonùv algoritmus má ov¹em znaèné nevýhody. Funguje pouze pro~racionální kapacity a~je pomìrnì pomalý. Nyní si~uká¾eme jiný algoritmus, který nevylep¹uje tok pomocí cest, ale pomocí tokù\dots Budeme k~tomu potøebovat sí» rezerv.
-
-\s{Definice:} {\I Sí» rezerv} k~toku~$f$ v~síti $S=(V,E,z,s,c)$ je sí» $R=(S,f)=(V,E,z,s,r)$, kde~$r(e)$ je rezerva hrany~$e$ v~toku~$f$.
-
-\s{Konvence:} Pro~hranu~$e$ znaèí~$\overleftarrow{e}$ hranu opaènou. Napø. pokud $e=uv$, tak $\overleftarrow{e}=vu$.
-
-Je dùle¾ité si~uvìdomit, ¾e sí» rezerv je závislá jak na~pùvodní síti~$S$, tak na~nìjakém toku~$f$ v~síti~$S$. Sí» rezerv~$R$ se~pak od sítì~$S$ li¹í pouze kapacitami -- sí»~$R$ má jako kapacitu hrany rezervu hrany v pùvodní síti. Pro~pøipomenutí: rezervu hrany~$e$ v~síti $S=(V,E,z,s,c)$ s~tokem~$f$ jsme si~definovali jako $r(e)=c(e) - f(e) + f(\overleftarrow{e})$.
-
-Ne¾ si~uká¾eme samotný algoritmus, doká¾eme si~následující lemma.
-
-\s{Lemma:} Pro~ka¾dý tok~$f$ v~síti~$S$ a~pro~ka¾dý tok~$g$ v~síti $R=(S,f)$ lze v~èase $\O(m)$ nalézt tok~$f'$ v~síti~$S$ takový, ¾e $\vert f' \vert = \vert f \vert + \vert g \vert$.
-
-\proof
-
-Dùkaz rozdìlíme do~tøí krokù. V~prvním kroku si~uká¾eme, jak budeme tok~$f'$ v~síti~$S$ konstruovat. V~druhém kroku doká¾eme, ¾e takto zkonstruované~$f'$ je opravdu tok. A~nakonec uká¾eme, ¾e splòuje po¾adovanou vlastnost, tedy ¾e jeho velikost je souèet velikostí tokù~$f$ a~$g$.
-
-\>{\it 1. konstrukce~$f'$}
-
-\noindent
-Pro ka¾dou dvojici hran~$e, \overleftarrow{e}$ urèíme~$f'(e)$ a~$f'(\overleftarrow{e})$ následovnì:
-
-\itemize\ibull
-\:Pokud~$g(e) = g(\overleftarrow{e}) = 0$, pak nastavme:
- \itemize\ibull
- \:$f'(e) := f(e)$,
- \:$f'(\overleftarrow{e}) := f(\overleftarrow{e})$.
- \endlist
-
-\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) = 0$, pak polo¾me:
- \itemize\ibull
- \:$\varepsilon := \min (g(e), f(\overleftarrow{e}))$,
- \:$f'(e) := f(e) + g(e) - \varepsilon$,
- \:$f'(\overleftarrow{e}) := f(\overleftarrow{e}) - \varepsilon$.
- \endlist
-
-\:Pøípad~$g(e) = 0$ a~$g(\overleftarrow{e}) > 0$ vyøe¹íme obdobnì.
-
-\:Pokud~$g(e) > 0$ a~$g(\overleftarrow{e}) > 0$, pak odeèteme od toku~$g$ cirkulaci po cyklu tvoøeném hranami~$e$ a $\overleftarrow{e}$:
- \itemize\ibull
- \:$\delta := \min (g(e),g(\overleftarrow{e}))$,
- \:$g'(e) := g(e) - \delta$,
- \:$g'(\overleftarrow{e}) := g(\overleftarrow{e}) - \delta$.
- \endlist
-
- Tok $g'$ nyní spadá pod nìkterý z~pøedchozích pøípadù, které u¾ umíme vyøe¹it.
-
-\endlist
-
-\>{\it 2. $f'$ je tok}
-
-\numlist{\ndotted}
-
-\:Nejdøíve ovìøme první podmínku: $\forall e \in E: 0 \leq f(e) \leq c(e)$. Vezmìme libovolnou hranu~$e \in E$. Podle toho, co teèe po~hranách~$e$ a~$\overleftarrow{e}$ v~toku~$g$, jsme rozdìlili konstrukci toku na~tøi pøípady:
-
- \numlist{\ndotted}
-
- \:Pokud po~hranách~$e$ a~$\overleftarrow{e}$ netekl ¾ádný tok~$g$, pak jsme nastavili $f'(e) := f(e)$ a~$f'(\overleftarrow{e}) := f(\overleftarrow{e})$. Tedy pokud~$f$ dodr¾oval kapacity, tak pro~$f'$ musí platit to samé.
-
-
- \:Pokud po~hranì~$e$ tekl tok~$g$ nenulový a~po opaèné nulový, tak jsme zvolili: $f'(e) := f(e) + g(e) - \varepsilon$. Víme, ¾e jsme si~$\varepsilon$ vybrali tak, ¾e $\varepsilon \leq g(e)$. Proto $f'(e) \geq 0$.
-
- Teï ovìøme, ¾e $f'(e) \leq c(e)$. V~pøípadì, ¾e $\varepsilon = g(e)$, tak $f'(e) = f(e) \leq c(e)$. V~opaèném pøípadì platí, ¾e $\varepsilon = f(\overleftarrow{e})$. Pak ov¹em
- $$f'(e) = f(e) + g(e) - f(\overleftarrow{e}) \leq $$
- $$\leq f(e) + \left[ c(e) - f(e) + f(\overleftarrow{e}) \right] - f(\overleftarrow{e}) = c(e).$$
- Vyu¾ili jsme, ¾e~$g$ je tok v~síti rezerv, tedy $g(e) \leq c(e) - f(e) + f(\overleftarrow{e})$.
-
- Pro tok $f'(\overleftarrow{e})$ platí, ¾e $\varepsilon \leq f(\overleftarrow{e})$. Proto $f'(\overleftarrow{e}) = f(\overleftarrow{e}) - \varepsilon \geq 0$. Zároveò $f'(\overleftarrow{e}) \leq f(\overleftarrow{e}) \leq c(\overleftarrow{e})$.
-
- Tím jsme dokázali, ¾e~$f'(e)$ i~$f'(\overleftarrow{e})$ dodr¾ují kapacity.
-
- \:V posledním pøípadì tekl po~obou hranách kladný tok~$g$. Men¹í tok z~$g(e)$ a~$g(\overleftarrow{e})$ jsme vynulovali a~od vìt¹ího odeèetli ten men¹í. Tok~$g'(e)$ a~$g'(\overleftarrow{e})$ tedy zùstal korektní a~tok~$f'$ u¾ konstruujeme podle pøedchozího pøípadu.
-
- \endlist
-
-\:Teï musíme je¹tì dokázat, ¾e nový tok neporu¹uje Kirchhoffovy zákony: $$\forall v~\in V \setminus \{z,s\}: f'^\Delta(v)=0.$$
- % neboli $$\forall v~\in V \setminus \{z,s\}: \sum_{u: uv \in E}{f'(uv)}=\sum_{u: vu \in E}{f'(vu)}.$$
-
- Vezmìme si~libovolnou hranu~$e = uv \in E$. Uvìdomme si, ¾e pøi~pøechodu z~$f(e)$ na~$f'(e)$ a~z~$f(\overleftarrow{e})$ na~$f'(\overleftarrow{e})$ bylo:
- \itemize\idot
- \:$f^\Delta(u)$ sní¾eno o~$g(e)$
- \:$f^\Delta(v)$ zvý¹eno o~$g(e)$.
- \endlist
- Seèteme-li úpravy na v¹ech hranách, dostaneme: $$f'^\Delta(v) = f^\Delta(v) + \sum_{u:uv \in E} g(uv) - \sum_{u:vu \in E} g(vu) =$$ $$= f^\Delta(v) + g^+(v) - g^-(v) = f^\Delta(v) + g^\Delta(v).$$
-
- Jeliko¾~$f$ byl tok, tak $f^\Delta(v) = 0$ a jeliko¾~$g$ byl tok, tak $g^\Delta(v) = 0$. Proto $f'^\Delta(v) = f^\Delta(v) + g^\Delta(v) = 0$.
-
-\endlist
-
-Tím jsme dokázali, ¾e~$f'$ je tok v~síti~$S$.
-
-\>{\it 3. $\vert f' \vert = \vert f \vert + \vert g \vert$}
-
-Pou¾ijme vztah pro souèet pøebytkù z pøedchozího kroku:
-$$\vert f' \vert = f'^\Delta(s) = f^\Delta(s) + g^\Delta(s) = \vert f \vert + \vert g \vert.$$
-\qed
-
-
-Pro algoritmus budeme potøebovat vybírat kvalitní toky~$g$ v~síti rezerv. Pokud se~nám to bude daøit, bude se~tok~$f'$ rychle zvìt¹ovat, a¾ bychom mohli dojít k~maximálnímu toku. Nejlépe by se~nám hodily co nejvìt¹í toky v~síti rezerv. Kdybychom si~dali za cíl najít v¾dy maximální tok v~síti rezerv, výsledek by byl sice krásný (dostali bychom tak rovnou i~maximální tok v~pùvodní síti), ale problém hledání maximálního toku bychom pouze pøenesli na~jinou sí». Na¹e po¾adavky na~tento tok budou tedy takové, aby byl dostateènì velký, ale abychom bìhem jeho hledání nestrávili moc èasu. Podívejme se, jak se~s~tímto problémem vyrovná {\I Dinicùv algoritmus}. Nejdøíve si~ale zadefinujme nìkolik pojmù.
-
-\s{Definice:} Tok~$f$ je {\I blokující}, jestli¾e pro~ka¾dou orientovanou cestu~$P$ ze~$z$ do~$s$ existuje hrana~$e \in P$ taková, ¾e $f(e) = c(e)$.
-
-\s{Definice:} Sí» je {\I vrstevnatá (proèi¹tìná)}, kdy¾ v¹echny vrcholy a~hrany le¾í na~nejkrat¹ích cestách ze~$z$ do~$s$.
-
-Dinicùv algoritmus zaèíná s~nulovým tokem. Potom v¾dy podle toku~$f$ sestrojí sí» rezerv a~v~ní vyma¾e hrany s~nulovou rezervou. Pokud v~této promazané síti rezerv neexistuje cesta ze~zdroje do~stoku, tak skonèí a~prohlásí tok~$f$ za maximální. Jinak proèistí sí» rezerv tak, aby se~z ní stala vrstevnatá sí» (rozdìlí vrcholy do~vrstev podle vzdálenosti od zdroje a~odstraní pøebyteèné hrany). Ve~vrstevnaté síti najde blokující tok, pomocí nìho¾ zlep¹í tok~$f$. Pak opìt pokraèuje sestrojením sítì rezerv na~tomto vylep¹eném toku~$f$ atd.
-
-\figure{dinic-cistasit.eps}{Proèi¹tìná sí» rozdìlená do~vrstev}{0.4\hsize}
-
-\s{Algoritmus (Dinicùv)}
-
-\algo
-\:$f \leftarrow$ nulový tok.
-\:Sestrojíme sí» rezerv~$R$ a~sma¾eme $e: r(e) = 0$.
-\:$l \leftarrow$ délka nejkrat¹í cesty ze~$z$ do~$s$ v~$R$.
-\:Pokud $l = \infty$, zastavíme se~a vrátíme~$f$.
-\:Proèistíme sí» $R \rightarrow$ sí»~$C$.
-\:$g \leftarrow$ blokující tok v~$C$.
-\:Zlep¹íme tok~$f$ pomocí~$g$.
-\:GOTO 2.
-\endalgo
-
-\s{Pozorování:} Pokud se~algoritmus zastaví, vydá maximální tok.
-
-\proof
-Víme, ¾e~$f$ je stále korektní tok (jediné, jak ho mìníme je pøièítání toku~$g$, co¾ je, jak jsme si~dokázali, \uv{ne¹kodná operace}). Jakmile neexistuje cesta ze~$z$ do~$s$ v~$R$, tak je $f$ maximální tok, nebo» v~tuto dobu by se~zastavil (a vydal maximální tok) i~Fordùv-Fulkersonùv algoritmus, který je korektní.
-\qed
-
-A teï je¹tì musíme ujasnit, jak budeme èistit sí» rezerv a~vybírat blokující tok~$g$.
-
-\s{Algoritmus proèi¹tìní sítì rezerv}
-
-\algo
-\:Rozdìlíme vrcholy do~vrstev podle vzdálenosti od~$z$.
-\:Odstraníme vrstvy za~$s$ (tedy vrcholy, které jsou od~$z$ vzdálenìj¹í ne¾~$s$), hrany do~minulých vrstev a~hrany uvnitø vrstev.
-\:Odstraníme \uv{slepé ulièky}, tedy vrcholy s~$\deg^{out}(v) = 0$, a~to opakovanì pomocí fronty. (Nejdøíve zaøadíme do~fronty v¹echny vrcholy s~ $\deg^{out}(v) = 0$. Pak dokud není fronta prázdná, v¾dy vybereme vrchol~$v$ z~fronty, odstraníme~$v$ a~v¹echny hrany~$uv$. Pro~ka¾dý takový vrchol~$u$ zkontrolujeme, zda se~tím nesní¾il výstupní stupeò vrcholu~$u$ na~nulu ($\deg^{out}(u) = 0$). Pokud sní¾il, tak ho zaøadíme do~fronty.)
-\endalgo
-
-\figure{dinic-neprocistenasit.eps}{Neproèi¹tìná sí». Obsahuje zpìtné hrany, hrany uvnitø vrstvy a~slepé ulièky.}{0.45\hsize}
-
-Hledání blokujícího toku zaèneme s~tokem nulovým. Pak vezmeme v¾dy orientovanou cestu ze~zdroje do~stoku v~síti~$C$. V~této cestì najdeme hranu s~nejni¾¹í hodnotou výrazu $r(e) - g(e)$ (neboli $c(e) - f(e)$ v~pùvodní síti). Tuto hodnotu oznaèíme~$\varepsilon$. Pak ke~ v¹em hranám na~této cestì pøièteme~$\varepsilon$. Pokud tok~$g$ na~nìjaké hranì dosáhne kapacity hrany, co¾ je zde~$r(e)$, tak hranu vyma¾eme. Následnì sí» doèistíme, aby splòovala podmínky vrstevnaté sítì. A~pokud je¹tì existuje nìjaká orientovaná cesta ze~zdroje do~stoku, tak opìt pokraèujeme s~touto cestou.
-
-\s{Algoritmus hledání blokujícího toku}
-
-\algo
-\:$g \leftarrow$ nulový tok.
-\:Dokud existuje orientovaná cesta~$P$ ze~$z$ do~$s$ v~$C$, opakuj:
-\::$\varepsilon \leftarrow \min_{e \in P} (r(e) - g(e))$.
-\::Pro~$\forall e \in P: g(e) \leftarrow g(e) + \varepsilon$.
-\:::Pokud $g(e) = r(e)$, sma¾eme~$e$ z~$C$.
-\::Doèistíme sí» zase pomocí fronty.
-\endalgo
-
-\s{Èasová slo¾itost} Rozeberme si~jednotlivé kroky algoritmu.
-
-\numlist{\ndotted}
-\:Inicializace toku~$f$ \dots $\O(m)$.
-\:Sestrojení sítì rezerv a~smazání hran s~nulovou rezervou \dots $\O(m + n)$.
-\:Najití nejkrat¹í cesty (prohledáváním do~¹íøky) \dots $\O(m + n)$.
-\:Zkontrolování délky nejkrat¹í cesty \dots $\O(1)$.
-\:Proèi¹tìní sítì \dots $\O(m + n)$.
- \numlist{\ndotted}
- \:Rozdìlení vrcholù do~vrstev -- provedlo ji¾ prohledávání do~¹íøky \dots $\O(1)$.
- \:Odstranìní nìkterých hran \dots $\O(m + n)$.
- \:Odstranìní \uv{slepých ulièek} pomocí fronty -- ka¾dou hranu odstraníme nejvý¹e jedenkrát, ka¾dý vrchol se~dostane do~fronty nejvý¹e jedenkrát \dots $\O(m + n)$.
- \endlist
-\:Najití blokujícího toku~$g$ \dots $\O(m \cdot n)$.
- \numlist{\ndotted}
- \:Inicializace toku~$g$ \dots $\O(m)$.
- \:Najití orientované cesty v~proèi¹tìné síti rezerv (staèí vzít libovolnou cestu ze~zdroje, nebo» ka¾dá z~nich v~této síti vede do~stoku) \dots $\O(n)$.
- \:Výbìr minima z~výrazu $r(e) - g(e)$ pøes v¹echny hrany cesty -- ta mù¾e být dlouhá nejvý¹e~$n$ \dots $\O(n)$.
- \:Pøepoèítání v¹ech hran cesty \dots $\O(n)$.
- \:Smazání hran cesty, jejich¾ tok~$g(e)$ se~zvý¹il na~hodnotu~$r(e)$ \dots $\O(n)$.
- \:Doèi¹»ování vyøe¹me zvlá¹».
- \endlist
-
- Vnitøní cyklus (kroky 2 a¾ 6) provedeme nejvý¹e~$m$ krát, nebo» pøi~ka¾dém prùchodu vyma¾eme alespoò jednu hranu (tak jsme si~volili~$\varepsilon$).
-
- Èi¹tìní bìhem celého hledání blokujícího toku~$g$ v~proèi¹tìné síti rezerv trvá dohromady $\O(m + n)$, nebo» ka¾dou hranu a~vrchol sma¾eme nejvý¹e jedenkrát.
-
- Najití blokujícího toku bude tedy trvat $\O(m \cdot n + (m + n)) = \O(m \cdot n)$.
-
-\:Zlep¹ení toku~$f$ pomocí toku~$g$ \dots $\O(m)$.
-\:Skok na~2. krok \dots $\O(1)$.
-\endlist
-
-Zbývá nám jen urèit, kolikrát projdeme vnìj¹ím cyklem (fází). Doká¾eme si~lemma, ¾e hodnota~$l$ vzroste mezi prùchody vnìj¹ím cyklem (fázemi) alespoò o~1. Z~toho plyne, ¾e vnìj¹ím cyklem mù¾eme projít nejvý¹e $n$-krát, nebo» cesta v síti na~$n$ vrcholech mù¾e být dlouhá nejvý¹e $n$.
-
-Uvìdomme si, ¾e uvnitø vnìj¹ího cyklu pøevládá èlen $\O(m \cdot n)$, tak¾e celková èasová slo¾itost bude $\O(n^2 \cdot m)$.
-
-\s{Lemma:} Hodnota~$l$ (délka nejkrat¹í cesty ze~$z$ do~$s$ v~proèi¹tìné síti) vzroste mezi fázemi alespoò o~1.
-
-\proof
-
-Podíváme se~na~prùbìh jednoho prùchodu vnìj¹ím cyklem. Délku aktuálnì nejkrat¹í cesty ze~zdroje do~stoku oznaème~$l$. V¹echny pùvodní cesty délky~$l$ se~bìhem prùchodu zaruèenì nasytí, proto¾e tok~$g$ je blokující. Musíme v¹ak dokázat, ¾e nemohou vzniknout ¾ádné nové cesty délky~$l$ nebo men¹í. V~síti rezerv toti¾ mohou hrany nejen
-ubývat, ale i~pøibývat: pokud po¹leme tok po~hranì, po~které je¹tì nic neteklo, tak v~protismìru z~dosud nulové rezervy vyrobíme nenulovou. Rozmysleme si~tedy, jaké hrany mohou pøibývat.
-
-Hrany mohou pøibývat jen tehdy, kdy¾ jsme po~opaèné hranì nìco poslali. Ale my nìco posíláme po~hranách pouze z~vrstvy do~té následující. Hrany tedy pøibývají do~minulé vrstvy.
-
-Vznikem nových hran by proto mohly vzniknout nové cesty ze~zdroje do~stoku, které pou¾ívají zpìtné hrany. Jen¾e cesta ze~zdroje do~stoku, která pou¾ije zpìtnou hranu, musí alespoò jednou skoèit o~vrstvu zpìt a~nikdy nemù¾e skoèit o~více ne¾ jednu vrstvu dopøedu, a~proto je její délka alespoò $l+2$. Pokud cesta novou zpìtnou hranu nepou¾ije, má buï délku~$> l$, co¾ je v~poøádku, nebo má délku~$= l$, pak je zablokovaná.
-
-Tím je lemma dokázáno.
-\qed
-
-\figure{dinic-cestashranouzpet.eps}{Cesta u¾ívající novou zpìtnou hranu}{0.4\hsize}
-
-V¹echna dokázaná tvrzení mù¾eme shrnout do~následující vìty:
-
-\s{Vìta:} Dinicùv algoritmus najde maximální tok v~èase $\O(n^2\cdot m)$.
-
-\s{Poznámka:} Algoritmus se~chová hezky na~sítích s~malými celoèíselnými kapacitami, ale kupodivu i~na~rùzných jiných sítích. Èasto se~pou¾ívá, nebo» se~chová efektivnì. A~je mnoho zpùsobù, jak ho je¹tì vylep¹ovat, èi odhadovat ni¾¹í slo¾itost na~speciálních sítích.
-
-\s{Poznámka:} Algoritmus nevy¾aduje racionální kapacity! Dal¹í z~dùvodù, proè maximální tok existuje i~v~síti s~iracionálními kapacitami.
-
-
-%k,s,v,na,do,ke,pro,pøi,a,u,i,po,
-
-\bye
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projects / ads2.git / commitdiff