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Si tous les nSuds n ont pas encore t visits, choisir un nSud non encore visit et recommencer 3. PqnPnPPfnPnPPq      f*   ZParcours en largeur  Placer le nSud de dpart dans la file Tant que la file n est pas vide soit n le premier nSud de la file traiter n enlever n de la file placer tous les nSuds adjacents n et non encore traits dans la file Fin tant queFZZ ZF% /$   [Parcours en largeur  .Une file est une structure de donnes de type FIFO (First In First Out) Enqueue(x,q) insre l item x la fin de q Dequeue(q) renvoie et enlve l item en tte de la file Empty(q) indique si la file est pleine, ne peut plus accepter d ajouts Initialize_queue(q) cre une file vide&HH  \Parcours en largeur] Exemple (BFS) ^ Exemple (BFS) _ Exemple (BFS) ` Exemple (BFS) a Exemple (BFS) b Exemple (BFS) c  Exemple (BFS) d! Exemple (BFS) e" Exemple (BFS) f#Parcours en largeur  PROCEDURE bfs(VAR g : graph; start : integer); VAR q : queue; { queue of vertices to visit } BEGIN FOR i := 0 TO g.nvertices-1 DO BEGIN processed[i] := false; discovered[i] := false; END; init_queue(q); enqueue(q, start); discovered[start] := true; P G Dfinitions   Chemin (de longueur n): Une squence de nSuds v0, v1, & , vn avec une arte de vi vi+1 pour chaque 0 <= i < n. Un chemin est simple si tous les nSuds dans le chemin sont distincts. Un cycle est un chemin de longueur 1 ou plus reliant un nSud vi lui mme. Un cycle est simple si le chemin est simple, sauf que les premiers et derniers nSuds sont les mme. ^rPFPLPdPP'      65  Ql {7Composantes connexes  Une composante connexe d un graphe non orient est l ensemble maximal de nSuds tel qu il existe un chemin entre toutes les paires de nSuds de cet ensemble Elles se dterminent par un parcours (ici DFS)  |8Composantes connexes   c := 0; FOR i := 0 TO g.nvertices-1 DO IF NOT discovered[i] THEN BEGIN c := c+1; write('Component ',c,': '); dfs(g,i); writeln END P }9Recherche de cycles  tUne arte arrire (u,v) est telle que v est un pre de u dans l arbre DFS construit pour un graphe. Toute arte arrire allant de u vers son pre v cre un cycle dans le chemin de v u.  ~:Recherche de cycles  PROCEDURE process_edge(x,y : integer); BEGIN IF parent[x] <> y THEN BEGIN write('Cycle from ',y, ' to ', x, ' : '); find_path(y,x); writeln { finished := true} END ENDZ*   ;Recherche de cycles  PROCEDURE find_path(start, endv : integer); BEGIN IF (start = endv) OR (endv = -1) THEN write(start:3) ELSE BEGIN find_path(start, parent[endv]); write(endv:3) END END;P v3Tri topologique  Le tri topologique d'un graphe orient sans circuit est une numrotation des sommets dans laquelle les descendants d'un sommet de numro k sont ncessairement de numro suprieur k. 6+ w4Topological Sort: DFSl)Tri topologique)Soit le graphe orient acyclique suivant )*(m*Tri topologiqueLa table indique pour chacun des nSuds (1 6) le nombre d artes entrantes (indegree) WXW n+Tri topologique y5Tri topologique  La valeur en tte de file est enleve (ici 1). La valeur indegree de chaque nSud adjacent au nSud enlev de la file est dcrmente de 1. Le(s) nSud(s) dont l indegree passe 0 sont mis dans la file (ici 2).69^*  o,Tri topologique  p-Tri topologique q.Tri topologique r/Tri topologique s0Tri topologique t1Tri topologique u2Tri topologiqueHLa file est vide est le rsultat du tri topologique est 1, 6, 2, 5, 3, 47z6Tri topologique   compute_indegrees(g, indegree); init_queue(zeroin); FOR i := 0 TO g.nvertices-1 DO IF indegree[i]=0 THEN enqueue(zeroin,i); j := 0; WHILE NOT empty(zeroin) DO BEGIN x := dequeue(zeroin); sorted[j] := x; j := j+1; FOR i := 0 TO g.degree[x]-1 DO BEGIN y := g.edges[x][i]; indegree[y] := indegree[y] - 1; IF indegree[y] = 0 THEN enqueue(zeroin,y) END END;P <0Bi-coloriage d un graphe  nonc: Ralisez un programme Pascal qui indique s il est possible de repeindre n gares d un rseau ferr de manire ce que deux gares relies directement ne soient jamais peintes de la mme couleur. On ne dispose que de deux couleurs. :H  =0Bi-coloriage d un graphe  @In 1976 the ``Four Color Map Theorem" was proven with the assistance of a computer. This theorem states that every map can be colored using only four colors, in such a way that no region is colored using the same color as a neighbor region. Here you are asked to solve a simpler similar problem. You have to decide whether a given arbitrary connected graph can be bicolored. That is, if one can assign colors (from a palette of two) to the nodes in such a way that no two adjacent nodes have the same color. To simplify the problem you can assume: no node will have an edge to itself. the graph is nondirected. That is, if a node a is said to be connected to a node b, then you must assume that b is connected to a. the graph will be strongly connected. That is, there will be at least one path from any node to any other node. %PPP%S#tA >0Bi-coloriage d un graphe  jInput The input consists of several test cases. 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H p2   H o?) U 2   Te o?9  :2/ p2   H o? R|B   T D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   He? 9u   Hxe? p  9v   HPe?87 9w   H(e?W w  9x   He?] m }  9y   Hزe?Z >z  9z B  Z Do?qN q|B  T D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B +T 0 {s (  ~  s *e U  e v2  N o?   H`e? F1/( 2  T8e o?'    ;4/5   He?= G]  D  2  Te o?# <  :3/ B  Z Do? H p2   H o?) U 2   Te o?9  :2/ p2   H o? R|B   T D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   Hz  9z B  Z Do?qN q|B  T D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B +U 0 |t (  ~  s *e U  e v2  N o?   He? F1/( 2  Te o?'    ;4/5   H e?= G]  D  2  Te o?# <  ;3/6 B  Z Do? H p2   H o?) U 2   Tg o?9  :2/ p2   H o? R|B   T D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   H g? 9u   Hg? p  9v   Hg?87 9w   Hg?W w  9x   Hg?] m }  9y   HPg?Z >z  9z B  Z Do?qN q|B  T D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B +V 0 }u (  ~  s *l#g U  g v2  N o?   HT%g? F1/( 2  T,*g o?'    ;4/5   H8.g?= G]  D  2  TP2g o?# <  ;3/6 B  Z Do? H p2   H o?) U 2   T@7g o?9  ;2/7 p2   H o? R|B   T D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   H?g? 9u   HFg? p  9v   HIg?87 9w   HHg?W w  9x   Hz  9z B  Z Do?qN q|B  T D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B +\ 0 {(  ~  s *]g U  g v2  N o?   HP`g? F1/( 2  T(eg o?'    ;4/5   HTAg?= G]  D  2  T Cg o?# <  ;3/6 B  Z Do? H p2   H o?) U 2   TYg o?9  ;2/7 p2   H o? RB   Z D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   HPrg? 9u   Hsg? p  9v   H g?87 9w   Hg?W w  9x   Hg?] m }  9y   Hg?Z >z  9z B  Z Do?qN q|B  T D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B +] 0 | (  ~  s *yg U  g v2  N o?   Hܔg? G1/8( 2  Tg o?'    ;4/5   H0xg?= G]  D  2  Tg o?# <  ;3/6 B  Z Do? H p2   H o?) U 2   T,g o?9  ;2/7 p2   H o? RB   Z D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   Hԑg? 9u   Hg? p  9v   HDg?87 9w   Hlg?W w  9x   H(g?] m }  9y   Hg?Z >z  9z B  Z Do?qN q|B  T D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B + 0 0h(  ~  s *\g U  g v2  N o?   Nlg? G1/8( 2  Tg o?'    ;4/5   Htg?= G]  D  2  Ttg o?# <  ;3/6 B  Z Do? H p2   H o?) U 2   Tg o?9  ;2/7 2   Tg o? R :9/ B   Z D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   HHg? 9u   HPg? p  9v   Hg?87 9w   Hg?W w  9x   Hg?] m }  9y   Hg?Z >z  9z B  Z Do?qN q|B  T D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B + 0 @h(  ~  s *h U  h v2  N o?   Hh? G1/8( 2  Td h o?'    ;4/5   Hph?= G]  D  2  TXh o?# <  ;3/6 B  Z Do? H p2   H o?) U 2   Th o?9  ;2/7 2   Th o? R :9/ B   Z D?LL* B   Z Do? 0 |B  T D?#6 B  Z D?T Q   Hh? 9u   H@#h? p  9v   H'h?87 9w   H+h?W w  9x   Hp/h?] m }  9y   H3h?Z >z  9z B  Z Do?qN qB  Z D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B + 0 -%P(  ~  s *;h U  h v2  N o?   Hg? G1/8( 2  T8Ah o?'    ;4/5   H8Fh?= G]  D  2  TIh o?# <  ;3/6 B  Z Do? H 2   TMh o?) U  ;10/ 2   TxLh o?9  ;2/7 2   TGh o? R :9/ B   Z D?LL* B   Z Do? 0 B  Z Do?#6 B  Z D?T Q   H Zh? 9u   Ht=h? p  9v   Hah?87 9w   Heh?W w  9x   H`ih?] m }  9y   H8mh?Z >z  9z B  Z Do?qN qB  Z D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B +  0 3+`(  ~  s *z  9z B  Z Do?qN qB  Z D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B + 0 =5p(  ~  s *h U  h v2  N o?   Hh? G1/8( 2  Th o?'    ;4/5   HLh?= G]  D  2  Tg o?# <  ;3/6 B  Z Do? H 2   T h o?) U  K10/11* 2   Th o?9  ;2/7 2   Th o? R :9/ B   Z D?LL* B   Z Do? 0 B  Z Do?#6 B  Z D?T Q   Hh? 9u   Hh? p  9v   Hdh?87 9w   Hh?W w  9x   H|h?] m }  9y   Hh?Z >z  9z B  Z Do?qN qB  Z D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B + 0 ?7(  ~  s *h U  h v2  N o?   HPh? G1/8( 2  Th o?'    ;4/5   Hxh?= G]  D  2  Th o?# <  ;3/6 B  Z Do? H 2   Th o?) U  K10/11* 2   Tj o?9  ;2/7 2   Tj o? R <9/12 B   Z D?LL* B   Z Do? 0 B  Z Do?#6 B  Z D?T Q   H j? 9u   H@j? p  9v   Hj?87 9w   H\j?W w  9x   H4j?] m }  9y   Hj?Z >z  9z B  Z Do?qN qB  Z D? _   s 0e0e    ByCtDE@F  5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||,0Xbkt9ZY4ygP9X"a$$ @     X  H  0޽h ? 3f3y___PPT10Y+D=' = @B +  0 (  r  S $j U  j r  S %j   j    `(j ??p1 \$Extrait partiel du programme Pascal$  H  0޽h ? 333380___PPT10. 4  0 4(  r  S 8Dj U  j   S @.j   j 0H  0޽h ? 333380___PPT10.p }  0 $(  r  S ^j U  j r  S _j   j H  0޽h ? 3333___PPT10i.\%+D=' = @B +}  0 $(  r  S qj U  j r  S rj   j H  0޽h ? 3333___PPT10i. DM+D=' = @B + 0 $$%(  ~  s *@xj U  j p2  H o?%  H{j?7W F( v2  N o?!C   H j?Ie i E0( vB  N D? Op2  H o? C     Hj? |  F( p2   H o? \     H? $ !  F( vB   N D?S S p2   H o?    H?   F( vB  N D?Y S Y p2  H o?$ Y   Hf?6!  V F( p2  H o?*  Hj?<\ F( vB  N D?P p2  H o?    Hj?   F( vB  N D?SS vB  N D? vB  N D? vB  N D?BB vB  N D??{   Hpj?!A 9r   H|j?'k G 9s   Hj?-H M 9t   Hj?3S 9u    Hj?   9v  ! Hj?   9w  " Hj? E   9x  # Hj?   9y  $ Njo? $ M Q: s 0$  H  0޽h ? 3f3y___PPT10Y+D=' = @B + 0 A9$$(  ~  s *dj U  j v2  N o?%  Htj?Mm E1( v2  N o?!C   Hj?Ie i E0( |B  T Do? Ov2  N o? C     Hlj? e  E1( p2   H o? \     Hj? $ !  F( vB   N D?S S p2   H o?    Hj?   F( vB  N D?Y S Y p2  H o?$ Y   Hj?6!  V F( p2  H o?*  Hj?<\ F( vB  N D?P p2  H o?    H0j?   F( vB  N D?SS |B  T Do? vB  N D? vB  N D?BB vB  N D??{   Hj?!A 9r   Hxj?'k G 9s   HPj?-H M 9t   HDm?3S 9u    H8m?   9v  ! Hm?   9w  " H m? E   9x  # Hxm?   9y  $ N mo? $ TQ: w r 1 1$ H  0޽h ? 3f3y___PPT10Y+D=' = @B +6 0 ]U$$(  ~  s *m U  m v2  N o?%  H m?Mm E1( v2  N o?!C   H$m?Ie i E0( |B  T Do? Ov2  N o? C     H4*m? e  E1( v2   N o? \     H0m? 6  E2( |B   T Do?S S p2   H o?    HH3m?   F( vB  N D?Y S Y v2  N o?$ Y   H8m?:3  Z E2( p2  H o?*  H =m?<\ F( vB  N D?P p2  H o?    HBm?   F( vB  N D?SS |B  T Do? vB  N D? vB  N D?BB |B  T Do??{   HHm?!A 9r   HNm?'k G 9s   H8Pm?-H M 9t   HTm?3S 9u    HhXm?   9v  ! H[m?   9w  " H_m? E   9x  # H$cm?   9y  $ Nfmo? > $ ZQ: r t x 1 2 2$ H  0޽h ? 3f3y___PPT10Y+D=' = @B +@ 0 g_$$(  ~  s *Dnm U  m v2  N o?%  H $ YQ: t x v 2 2 2$ H  0޽h ? 3f3y___PPT10Y+D=' = @B +Q 0 xp$$(  ~  s *m U  m v2  N o?%  Hm?Mm E1( v2  N o?!C   Hm?Ie i E0( |B  T Do? Ov2  N o? C     Hm? e  E1( v2   N o? \     H,m? 6  E2( |B   T Do?S S p2   H o?    H8m?   F( vB  N D?Y S Y v2  N o?$ Y   H\m?:3  Z E2( v2  N o?*  Nm?Ii E3( |B  T Do?P v2  N o?    Hm?   E2( |B  T Do?SS |B  T Do? vB  N D? vB  N D?BB |B  T Do??{   Hm?!A 9r   Hm?'k G 9s   Hm?-H M 9t   Hm?3S 9u    Hlo?   9v  ! Ho?   9w  " H o? E   9x  # Ho?   9y  $ Noo? S $ YQ: x v u 2 2 3$ H  0޽h ? 3f3y___PPT10Y+D=' = @B +h 0  $$(  ~  s *o U  o v2  N o?%  Ho?Mm E1( v2  N o?!C   H%o?Ie i E0( |B  T Do? Ov2  N o? C     No? e  E1( v2   N o? \     HT-o? 6  E2( |B   T Do?S S v2   N o?    HP5o?   E3( |B  T Do?Y S Y v2  N o?$ Y   HH 0 e]`""(  ~  s *Vp U  p v2  N o?%  HxZp?Mm E1( v2  N o?!C   H p?Ie i E0( |B  T Do? Ov2  N o? C     Hhsp? e  E1( v2   N o? \     H0xp? 6  E2( |B   T Do?S S v2   N o?    Hp?   E3( |B  T Do?Y S Y v2  N o?$ Y   H,p?:3  Z E2( v2  N o?*  Np?Ii E3( |B  T Do?P v2  N o?    Hp?   E2( |B  T Do?SS |B  T Do? |B  T Do??{   Hp?!A 9r   H8p?'k G 9s   Hdp?-H M 9t   Hp?3S 9u   Hp?   9v   Hp?   9w    Hp? E   9x  ! Hpp?   9y  " HHp?_ [ s  =BF Tree H  0޽h ? 3f3y___PPT10Y+D=' = @B +^  0 p ^(   r   S p U  p r   S (p m:  p >   0`p  +  WHILE NOT empty(q) DO BEGIN v := dequeue(q); process_vertex(v); processed[v] := true; FOR i := 0 TO g.degree[v]-1 DO BEGIN IF NOT discovered[g.edges[v][i]] THEN BEGIN enqueue(q,g.edges[v][i]); discovered[g.edges[v][i]] := true; END; IF NOT processed[g.edges[v][i]] THEN process_edge(v,g.edges[v][i]) END END END;P     `Tp ??f G l \$Extrait partiel du programme Pascal$  H   0޽h ? 333380___PPT10.;}  0 $(  r  S p U  p r  S `p   p H  0޽h ? 3333___PPT10i.p+D=' = @B +$  0 `$(  `r ` S 8p U  p r ` S p   p H ` 0޽h ? 333380___PPT10.?Ց$  0 d$(  dr d S lp U  p r d S 0p   p H d 0޽h ? 333380___PPT10.?$  0 h$(  hr h S $ u U  u r h S  u   u H h 0޽h ? 333380___PPT10.?xq$  0 l$(  lr l S @u U  u r l S u   u H l 0޽h ? 333380___PPT10.JP$  0 p$(  pr p S *u U  u r p S +u   u H p 0޽h ? 333380___PPT10.KpT$  0 pL$(  Lr L S ;u U  u r L S   ' 2 P  f\fugֳgֳ?Z,$D0 =H  C P # lgֳgֳ?BZL P # lgֳgֳ?Z Q Q   P  f@mugֳgֳ >?"W V___PPTMAC110(   hnamd` Arial&Monotype Typography    hnamd` Arial&Monotype Typography 4___PPT10 B___PPT9$ &Un ordre topologique: C G B A D E F H,' C P T$tu ??I= C W8 2   P Txu ??p pj  W3 2   P T4}u ??pj  W2 2   P TLu ??t) W1 2   P TDu ??f tW` W7 2   P Tu ??s m  W6 2   P Txu ?? x  W5 2   P Tu ??p j  W4 2  H P 0޽h ?`PPPPPPP P PPP P P PP PPPP PP P PP PPP PPP P !fffy___PPT10Y+D=' = @B +  0 $(  $~ $ s *Du U  u ~ $ s *u   u J $ C "A ts02` H $ 0޽h ? 33___PPT10i.BPD+D=' = @B +  0 (D(  (~ ( s *(u U  u ~ ( s *u   u J ( C "A ts02% g  ~ pp  (# #"*VUWVUV ( <u?@ p  I0 @` ( <Du? @ p  I6 @` ( <u?@    I2 @`  ( <|?@ p M3 @` 8 <G|?@  I0 @` 8 <O|?@  I2 @` 8 <8X|?@ p I0 @` 8 <`|?p@  I1 @``B 8 0o ?ppZB 8 s *1 ?ZB 8 s *1 ?ZB 8 s *1 ?ppZB 8 s *1 ?  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