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城市交通拥挤问题的探讨_英文_

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城市交通拥挤问题的探讨_英文_城市交通拥挤问题的探讨_英文_ M A T H EM A T ICA A P PL ICA TA () 2007, 20 1: 31, 36 Ξ A D iscu ss ion of the Tra f f ic Jam Prob lem in C it ie s () ) )((X IN Yu 2m e i 辛玉梅, RU A N R u 2jiang 阮如江, L E I H o ng2ch uan 雷洪川 (), , 200240, D ep a tm en t of M a th em a t icsS a ...
城市交通拥挤问题的探讨_英文_
城市交通拥挤问题的探讨_英文_ M A T H EM A T ICA A P PL ICA TA () 2007, 20 1: 31, 36 Ξ A D iscu ss ion of the Tra f f ic Jam Prob lem in C it ie s () ) )((X IN Yu 2m e i 辛玉梅, RU A N R u 2jiang 阮如江, L E I H o ng2ch uan 雷洪川 (), , 200240, D ep a tm en t of M a th em a t icsS a n g h a i J ia oton g U n iv e rs ity S a n g h a i C h in a rhh .A bstra c t: T h is p ap e r d iscu sse s th e m e tho d o f so lv ing th e p ro b lem o f t raff ic jam in c it ie s , A cco rd ing to th e sta tu s o f jam w e in t ro duce a co ncep t ca lled exp ed iencym ak ing th e sta tu s o f , , exp ed iency m ea su rab lem enaw h ilew e add th e exp ed iency fac to r in to th e p h y sica l d istance and ge t .a new k ind o f d istance so th a t th e a lgo r ithm sea rch ing fo r sho r te st p a th can st ill be su itab le ( ) B e side s th is p ap e r ta lk s abo u t th e sto rage m e tho d s: Coo rd ina te S to rage COO , Com p re ssed () () , , , Sp a r se R ow C SR Com p re ssed Sp a r se Co lum n C SC B lo ck Sp a r se R ow and th e sho r te st p a th : 2, a lgo r ithm D ijk st ra a lgo r ithm and B e llm anfo rd a lgo r ithm th en w e g ive th e step s o f d ijk st ra 2.a lgo r ithm and it s la te st sp eedup a lgo r ithm : ; ; Key word sT raff ic jam E xp ed ien cySho r te st p a th () AM S 2000Subjec t C la ss if ica t ion: 54C 60 CL C Num ber: O 189. 1 () : : 100129847 20070120031205D ocum en t codeA A r t ic le ID 1. In troduc t ion , A cco rd in g to th e da ta o f t raff ic re lea sed b y N a t io n a l B u reau o f S ta t ist ic s o f C h in asin ce , th e op en an d refo rm C h in e se eco nom y h a s deve lop ed ve ry rap id ly an d th e n um b e r o f veh ic le . , . h a s g row n ve ry qu ick lyH ow eve rth e co n st ru c t io n o f ro ad s h a s no t fo llow ed th e stepC it ie s’ , 1760002001, co n d it io n o f t raff ic is even w o r seth e ro ad’ s len g th w a s km in an d it cam e to 1910002002, 8. 5%. , 18. 22 km in in c rea sed b y H ow eve rth e re w e re m illio n c iv il veh ic le s in 2001, 20. 53 2002, 12. 7%. an d it cam e to m illio n in in c rea sed b y T h e len g th o f ro ad ow n ed b y 1 4. 92001, 5. 42002, 10. 2% , c it izen s o n ave rage w a s km in an d it cam e to km in in c rea sed b y . b u t it st ill co u ld no t ca tch up w ith th e g row th o f veh ic le sIn fac t som e m e t ropo lise s su ffe r f rom m o re seve re t raff ic jam s p ro b lem s. T ak e Sh an gh a i fo r ex am p le , th e len g th o f ro ad 2002 , 2003 , 3. 1 % , w a s 6286 km in an d 6484 km in in c rea sed b y b u t th e n um b e r o f 2 1. 39 2002 1. 74 2003 , 25 % . veh ic le s w a s m illio n in an d m illio n in in c rea sed b y B ecau se , , .it is pop u lo u s in c it ie sw ith veh ic le s p ro life ra tedth e t raff ic jam b ecom e s e sp ic ia lly seve re T h e m ism a tch o f supp ly an d dem an d o f t raff ic cau se s th e t raff ic jam an d it b ecom e s a Ξ Rece ived da te: F eb rua ry 23, 2006 : 2, , , , , B iographyX IN Yu m e ifem a leH an H e ilo ng jianga sso c ia te p ro fe sso rm a jo r in gene ra l topo lo gy and .a lgeb ra ic topo lo gy 2007 M A T H EM A T ICA A P PL ICA TA 32 t ro u b le som e p ro b lem to go ve rnm en t an d p eop le. , B ecau se th e co n st ru c t io n o f ro ad is a lo n g te rm in ve stm en tth is t raff ic p ro b lem can ’ t b e . so lved b y co n st ru c t io n in a sho r t t im eT h en how to u se th e lim ited ro ad re so u rce s an d to .choo se th e p rop e r p lace to co n st ru c t ro ad s b ecom e e sp ec ia lly im po r tan t ( ) In recen t yea r s, th e IT S’ In te lligen t T raff ic Sy stem re sea rch is b loom in g. T h e t raff ic cen te r o f m an agem en t t r ied to m o n ito r an d co n t ro l th e t raff ic sta tu s sim u ltan eo u sly to so lve . th e t raff ic p ro b lem an d to m ak e st ra igh tw ayT h is p ap e r in t ro du ce s th e co n cep t o f , . exp ed ien cyqu an t ita te s th e sta tu s o f th e t raff ic so th a t it can b e com p u ted b y com p u te r sM eanw h ile w e t ran sfo rm th e p h y sica l d istan ce to a v ir tu a l d istan ce in c lu d in g th e fac to r o f . . t im eT h is co u ld p ro v ide th e d r ive r s w ith th e qu e ry se rv ice o f th e sho r te st p a th sT h e . in fo rm a t io n is co llec ted b y th e de tec to r s in sta lled o n th e ro adD e tec to r s ge t t raff ic . , in fo rm a t io n an d sen d it b ack to th e co n t ro l cen te rA cco rd in g to th is in fo rm a t io n h e com p u te r , . se rve r ren ew s th e da tab a se to suppo r t qu e ry se rv iced irec t th e f low o f veh ic le sT h en ro ad . , re so u rce s can b e u sed effec t ive lyB e side s acco rd in g to lo n g te rm in fo rm a t io n w e can choo se .th e r igh t p lace to co n st ru c t n ew ro ad s 2. Pa th Conn ec tedn e ss X p , q X . p q X , Suppo se is a topo lo g ica l sp acea re po in t s in A p a th b e tw een an d in m ean s () () a co n t in uo u s m app in g Ρ?0, 1 ? X , Ρ0= p , Ρ 1= q. W e ca ll p an d q th e sta r t an d th e en d o f th e p a th Ρ. If b e tw een an y tw o po in t s in X , th e re is o n e p a th to co n n ec t th em , th en X .is ca lled p a th co n n ec ted sp ace 1 Suppo se X is a topo lo g ica l sp ace, p is a po in t in X , th en X is p a th co n n ec ted Theorem sp ace equ iva len t to an y q ? X , ex it o n e p a th to co n n ec t p an d q in X . . . x y Proof T h e n ece ssity is o b v io u slyFo llow in g is th e su ff ic ien cySuppo se an d a re tw o X , , f ,po in t s in acco rd in g to th e co n d it io n o f th e th eo rem th e re ex it tw o co n t in uo u s m app in g 1 f 2 () () f ?0, 1 ? X , f 0= p , f 1= x ; 1 1 1 f ?0, 12 () () ? X , f 2 0= p , f 2 1= y. C rea te a m app in g: t f 1 - , 1 Κ t ? 0, Κ, ( ) f ?0, 1 ? X , f t= (t - Κ t ? Κ, 1 , , f 2 1 - Κ 0 < Κ< 1 an d it is a co n stan t. () () () () () () () f 0= f 1= x , f 1= f 1= y , f Κ= f 0= f 0= p. f , f T h en 1 2 1 2 B ecau se 1 2 a re , f x y. x , co n t in uo u sso is co n t in uo u sth en th e re ex ist s a co n t in uo u s m app in g b e tw een an d So y . 1 .an d a re p a th co n n ec tedT h e T h eo rem is p ro ved Suppo se X is a topo lo g ica l sp ace, C < X , C is ca lled a p a th co n n ec ted b ifu rca t io n o f X , if C X is a p a th co n n ec ted su b se t o f an d it is no t th e p rop e r su b se t o f o th e r p a th co n n ec ted 3, 4X .su b se t o f X X 2 , Theorem Suppo se is a topo lo g ica l sp aceis a p a th co n n ec ted sp ace if an d o n ly if 2. : N o. 1 33 X IN Yu m e i e t a lA D iscu ssio n o f th e T raff ic J am P ro b lem in C it ie s X o n ly h a s o n e co n n ec ted b ifu rca t io n. 3. Ne twork an d Roa d Ne twork . N ew w o rk is a g rap h co n ta in in g som e ve r t ice s an d edge s b e tw een ve r t ice sG rap h ’ s :def in it io n is a s fo llow s v n }, E = {e1 , , em }, P = {p 1 , , Suppo se V = {v 1 , , p m } a re th ree f in ite se t s an d :sa t isfy ) 1V ;is no t em p ty () )2 fo r eve ry ei ? E , it is a d iso rde red o rde rede lem en t p a ir {v s , v t }; ) 3P an d E h ave th e sam e n um b e r o f e lem en t s an d h ave a b ijec t io n. () () V , E , P G , v e, T h en fo rm an u n d irec ted d irec ted edge w e igh t g rap h i is ca lled ve r tex i is , p e.ca lled edgei is ca lled i ’ s w e igh t () It is a ro ad w itho u t in n e r ve r tex , m a rk ed a s R. Supppo se V G = {v 1 , v 2 , , v n } is a () () E G = {e1 , e2 , , em } is a sim p le ro ad se t w ith th e po in t s o f V G a s ve r tex. ve r tex se t, () () () () ) () (N e tw o rk G is an o rde red fo u r e lem en t s g ro up V G , E G , ΥG ,
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