第 五 章 图 论 ( 第二部分 )
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第 五 章 图 论 ( 第二部分 ). 1. 通路 2. 图的 连通性 3. 赋权图的最短通路. 赋权图与边的权. [ 定义 ] 权 在图的点或者边上标明的信息 ( 数量 ) 称为 权。 边 e 的权记为 W(e) ； 如果用两个端点的序列 (u, v) 表示边，则边 (u,v) 的权记为 W(u,v) 。 规定： (1) W(uu) = 0 ， 若 (u, u) E(G), (2) W(uv) = ∞, 若 (u, v) E(G) 。 [ 定义 ] 赋权图 - PowerPoint PPT Presentation
Transcript of 第 五 章 图 论 ( 第二部分 )
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eW(e) (u, v)(u,v)W(u,v)
(1) W(uu) = 0 (u, u) E(G), (2) W(uv) = , (u, v) E(G)
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: W(a,b)=5W(a,a)=0W(b,d)=W(a,d)=8
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[]Guvuvuv
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AFBCDE
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acac(a, b, c)5+4=9;(a, c)12;(a, d, c)8+20= 28 abcac9
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(Dijkstra)G=(V,E)az (1) aV-{a}P1P1t1; t1 = zP1az (2)(2) a V-{a,t1}P2P2t2 t2 = zP2az (3)(3) aV-{a,t1,t2}P3P3t3 t3= zP3az (4)(4) kPkz
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G=(V, E)Gaz
[]T(1) T V(2) z T, z(3) a T, a
[] t1 Tat1Tt1TDT(t1).T={c, e, f, g, z}, DT(c).acT:(a, b, c): : 1+2 = 3(a, c): : 4(a, d, c): : 4+3 = 7 DT(c) = 3TG
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[]: T = {t1, t2, , tn}, t1TDT(t1) = min{DT(t1), DT(t2) , , DT(tn)} (1) at1DT(t1) (2) 2 in at1 atiT = {e, f, g, z}, DT( e ) = 9; DT( f ) = 6;DT( g ) = 8; DT( z ) = ; Taf afDT(f)=6G
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1 (1) () at1DT(t1). DT(t1)at1TDT(t1)T
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() dnew dnew < DT(t1) t2T dnew = DT(t2) + W(t2t1) > DT(t1) w(t2t1) t2t1 dnew < DT(t1) at1DT(t1).aTt1t2
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() (2) () ti(i2) atiat1Pdd
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G=(V, E)az 1T1V-{a}T1t1 t1=zDT1(t1)az 2 2T2=T1-{t1},T2t2 t2=z,DT2(t2)az (3) 3 TkzTk DTk(z)az
Ti
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Tm={tm, tm+1, , tn}DTm(ti)tiTtm Tm+1= Tm {tm}ti tmtntm+1tmTmTm+1tmtiW(tm,ti)
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G=(V, E)az
[](1) T = V {a}(2) vTDT(v)=W(a,v) (3) TtTTT{t} (4) t = zDT(z)az t z v TDT(v)=min(DT(v), DT(t)+W(t,v)) (3).
DT(z)az
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T1{b,c,d,e,f,g,z}: DT1(b) = 1 ab DT1(c) = 4 ac DT1(d) = 4 ad DT1(e) = DT1(f) = DT1(g) = DT1(z)=
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cT2abcadabeacabc,DT2(c)=3 DT1(b) = 1 ab DT1(c) = 4 ac DT1(d) = 4 ad DT1(e) = DT1(f) = DT1(g) = DT1(z)=
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adabceabcfabcgT3dDT3(d)=4adadad
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T4T3{d}={e, f, g, z}T4 abceabcfabcgT4fDT4(f)=6afabcfaf
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T4egDT5(e)=8aeabcfeae
T5T4{f}={e, g, z} T5 abcfeT5abcgabcg
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T6T5{e}={g, z} T6 T6T6fDT6(g)=6agabcgag
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T7T6{g}={z} T7 azDT7(z)=9 abcfezabcfezT7
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(1) T1=V {a} 1T1
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(2) T2=T1{b} T2t T23T2T1T1
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T3=T2{ c }T34T2T1T3
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ezfezcfezbcfezabcfez
bcdefgz44410968988109
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az
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: abcghzT1T2T3T4T5T6T7T8
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u0c
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Dijkstra u0edc
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P20335
DT(t1)at1at1dnew < DT(t1)
DT(t1)at1TDT(t1)T
t2Tdnew = DT(t2) + w(t2t1)
dnew < DT(t1)
DT(t2) < DT(t1)
at2at1 t1T DT(t1) = min{DT(t1), DT(t2) , , DT(tn)}DT(t1)at1at1dnew < DT(t1)
DT(t1)at1TDT(t1)T
t2Tdnew = DT(t2) + w(t2t1)
dnew < DT(t1)
DT(t2) < DT(t1)
at2at1 t1T DT(t1) = min{DT(t1), DT(t2) , , DT(tn)}DT(t1)at1at1dnew < DT(t1)
DT(t1)at1TDT(t1)T
t2Tdnew = DT(t2) + w(t2t1)
dnew < DT(t1)
DT(t2) < DT(t1)
at2at1 t1T DT(t1) = min{DT(t1), DT(t2) , , DT(tn)}
