第十章 图的概念与表示
DESCRIPTION
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 PresentationTRANSCRIPT
-
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{