第十章 图的概念与表示

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第十章 图的概念与表示. 10.1 图的基本概念 10.2 链 ( 或路 ) 与圈 ( 或回路 ) 10.3 图的矩阵表示. 退出. 10.1 图的基本概念. 什么是图 ? 可用一句话概括,即:图是用点和线来刻划离散事物集合中的每对事物间以某种方式相联系的数学模型。因为它显得太抽象,不便于理解,所以有必要给出另外的回答。下面便是把图作为代数结构的一个定义。. - PowerPoint PPT Presentation

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  • 10.1 10.2 ()()10.3

  • 10.1 ?

  • 10.1.1 GG=VEEV

  • GeEvivj(e)=vivjeeE(e)=evievje

  • vivj()evivj()evivjvi adj vjvi nadj vj()evivjvivj()e

  • GGG=VE10.1.2 G=GG

  • 10.1.3 G=()G()G

  • 10.1.4 G=G=W

  • 10.1.5 G=V(vi)(vj)(vivjVvivjE)K|V||V|=nKn

  • 10.1.6 G=vVvvd+(v)vvd-(v)vd(v)d(v)=d+(v)+d-(v)G=vVd(v)v2

  • G=(G)=max{d(v)|vV}(G)=min{d(v)|vV}G

  • 10.1.1 G=

    10.1.2

  • 10.1.7 G=k(v)(vVd(v)=k)GkkG(G)=(G)=k

  • 10.1.8 G1=G2=(1) V2V1E2E1G2G1G2G1(2) V2V1E2E1E2E1G2G1G2G1(3) V2=V1E2E1G2G1G2 G1

  • 10.1.9 G2=G1=uvuvE1uvE2G2V2G2V2GV2G2E2G2E2GE2

  • 10.1.10 G1=G2=G=E2=E-E1G2=G2GG1

  • 10.1.11 G=G1=E1E=G1GG1GG1= G

  • 10.1.12 ()G1=G2=fV2V1vuV1uvE1f(u)f(v)E2(E1E2)G1G2G1G2G1G2G2G1

  • 10.1.13

  • (1) (2) (3)

  • (a)3x1(b)3y1

    10.1.14(a)(b)

  • 10.1.13

  • 1.1.14

  • 10.2 ()()()()()()

  • 10.2.1 ()G=v0v1vmV()e1e2emEvi-1vieiv0e1v1e2v2emvmv0vm()v0vm()()()v0=vm()()

  • 10.2.2 ()()()()()()()()

  • 10.2.3 ()()()()()()()()

  • 10.2.1 vivj()vivj()10.2.2 n(1) ()n-1(2) ()n

  • 10.2.4 vivj()vivjvivjGGG(G)

  • 10.2.5 GGGG=SV-SESG-SS=|v|G-vG=TE-TTVG-TT={e}G-e

  • G=SVTESSVG+SS={v}G+vG=TETG+TTV

  • 10.2.6 G=SV(G-S)(G)TS(G-T)=(G)SGGS={v}vGGTGT={e}eG

  • G(G)= {|S||SG}GG(G)=0G(G)=1

  • (G)= {|T||TG}G(G)=0G(G)=0K1(K1)=0G(G)=1Kn(Kn)=n-1

  • (H.Whitney)193210.2.3 G(G)(G)(G)10.2.4 Gvuwuwv

  • 10.2.5 Geuwuwe10.2.6 Gee

  • GGG1G1G

  • 10.2.7 GGGGGGGG

  • 10.2.7 GGGGG1G1G10.2.8

  • 10.2.8 10.2.6{v1,v2,v3}

  • 10.2.9 10.2.10 uv()()()uvd

  • d0d=0d+dduvd=+uvdd

  • CPU

  • Ar1r2Br2r1

  • Pt={p1p2pm}tQtPttRt={r1r2rn}tGt=trii=12nEpkQtrirj

  • Rt={r1,r2,r3,r4}Qt={p1,p2,p3,p4}p1r4r1p2r1r2r3p3r2r3p4r3r1r4

  • Gt=10.2.7tGt10.2.7Gt

  • 10.2.7

  • 10.3 G=()

  • G=V

  • 10.3.1 G=V={v1v2vn}VnA=(aij)G

    ij=12n

    GGA(G)

  • GGHPA(H)=P-1A(G)PV

  • 1

  • AAGnn

  • nn

  • ivii1vij1vjd+(vi)= d-(vj)=

  • GAAlAlijalijGivijvjl10.3.1 AGAlijalijGvivjl()

  • vivjvivjGAA2A3AnAr 1vivjvivjAr

  • 10.2.2nn-1n 1rnBn=A+A2+A3++An

  • 10.3.2 G=V={v1v2vn}nP=(pij) 1 vivjPij= 0 PG

    {

  • ()()GAPBn=A+A2+A3++AnBn1P

  • BB={0,1}BC

  • BCBCBoC(BC)ij=bijcij(BoC)ij= (bikckj)i,j=1,2,,nbijcijBCij

  • AAoA=A(2)r=2,3,,A(r-1)oA=A(r)ArA(r)Ar vivjrA(r) vivj=1 =0PP=AA(2)A(3)A(n)= A(k)

  • G=EVVEB

  • X={x1,x2,,xm}Y={y1,y2,,yn}RXYMR=(rij)1, R rij= 0, i=1,2,,mj=1,2,,n{

  • RMRF(R)=V|V|=nRR+R+=RR2R3Rk (kn)R1R2A1A2R1R2A1A2R+ =

  • G=AEE2=EoEvkviEvkvkEvjviE2vjvivj2E2(i,j)1A(2)E2A(k)Ekk=2,3,,nA+=AA(2)A(3)A(n) A+=AA(2)A(3)A(n)=P

  • EE+A+A+PA(2)A(3)A(n) A(k)A+PWarshallAA+

  • (1) PA(2) k1(3) i1(4) pik=1j=1,2,,npijpijpkj(5) ii+1in(4)(6) kk+1kn(3)(4)pik=1iki

  • E1E2A=(aij)B=(bij)E1E2AB(AB)ij=aijbij10.3.2 GviP=(pij)GPT=(pji)PPPTi1vi

  • VP={p1,p2,,pn}PG=piVPi=1,2,,nEpipjGpipiGA=(aij)A+=( )A+ =1pi

  • G=W=(wij) w, w[vi,vj]E wij= 0 vivjWG{