8 顶点边网络的流 不到走迷重醛）=winttoa).ha)=min2,-在 (-2,1(m) 个寿检查过 ②，上9不海足起家件不到起进2 件里 .检查，继至，购上，a=4<e13=9,给起进为(+，》见过 minl(3.f4,3}=min2,1}=1m在上(，t)上t=c3,t=5,t不 ⑥.检查u,在至(，)：=8<t=10,故t 1() nia.eg-}=mm1,2=l,因平‘去到起 什柳足性 下一步限制过程件 (仁)、调 反霜宿踪，然超点零因一只起 足所示 净到一条必总链Q=购4，如图10-满 f2=f.2+0=5+1=6 1.3=1.3+0=4+1=5 f4-f4+-8+1-9 f12=12-0-4-1=3 f.3=f,3-9=1-1=0 见两上g任保特不大民制香去到达大图上一只零零可 总如图10-5所示 图10-5 人上述起进 起进在图0检查给，起进仁，卷拳 程寻 点链开标给起进(0，+四，检查，给 给你起进（+1,1)，检记.主（巴4，） f3=0. 上，人,均不符合起进条件起号过程无算进，算算结束故 图10-5给收零可 总意为该达武零最出总件最出总任为炉 (f)=f1+f2=ft+ft=14. 现已起进零超点集合巧=s,,未起进零超点集合了=小，故面 K(,了)={2，(，t}该达零最小有，两零容任c(,)-14 至由比可向，最小有零容任金小影最出总零值最小有达 月最紧张零那给 为提高最出启值必须多出有过零容程4是过来如果最外看过惑季这力被降低时 会最出零总降低件8 q❅r❇s✉t❇✈❅✇❇① min{+∞, 2} = 2✺ S ② ♦❅③❇❣✁❤✁◆✁★✁✧✺ (3). ④ ③✰✬✣✭✣⑤✣⑥✣❣✣❤✣★✣✧ v2, ❣✣❤ v2, ⑦❆ (v1, v2) ❢ ,f1,2 = 4 > 0, ❦✣❏ v1 ✬✣✭ (−v2, l(v1)), ❬✁⑧✁⑨:l(v1) = min{l(v2), f1,2} = min{2, 4} = 2✺⑩⑦❆ (v3, v2) ❢ ,f3,2 = 0, ❦ v3 ❧✁♠✁♥✬✁✭✺⑩⑦❆ (v2, v4) ❢ , f2,4 = c2,4 = 9, ♠✁❶✁❷✬✁✭✁❸✁❹,v4 ❺✁❧✁♠✁♥✬✁✭✺ v2 ❺ ② ♦❅③❇❣✁❤✁◆✁★✁✧✺ (4). ❣✣❤ v1, ⑦❆ (v1, v3) ❢ , f1,3 = 4 < c1,3 = 9, ❞ v3 ✬✣✭✣♦ (+v1, l(v3)), ✤✯♣, l(v3) = min{l(v1),(c1,3 − f1,3)} = min{2, 5} = 2✺ (5). ❣✁❤ v3, ⑦❆ (v4, v3) ❢ ,f4,3 = 1 > 0, ❦✁❞ v4 ✬✁✭✁♦ (−v3, l(v4)), ✤❅♣ l(v4) = min{l(v3), f4,3} = min{2, 1} = 1✺⑩⑦❆ (v3,t) ❢ ,f3,t = c3,t = 5,t ♠✁❻✬✁✭✺ (6). ❣✣❤ v4, ⑦❆ (v4,t) ❢ ,f4,t = 8 < c4,t = 10, ❦ t ✧❧✣♥✬✣✭ (+v4, l(t)), ✤✯♣ l(t) = min{l(v4),(c4,t − f4,t)} = min{1, 2} = 1, ❼✁❀ t ❧✁♥✬✁✭, ▲●✁❽✁✪✁❾✼✁✽◆✁❖✺ (❿)❜⑩➀✁➁✁➂✁➃ (1). “➄ ✮❇➅✁➆”, ❉✦✁✧✁★✁✩✁✪✁✫✁✬✁✭✱♥✪✁❸✳✁✴✁✵ Q = sv2v1v3v4t, ➇ ❚ 10–4 ✷ ✸✁❯✁❱✁✺ (2). ❉ θ = l(t) = 1 ✼✁✽✳✁✴✁✵❢✁➈✁❆✁★✴✾ : f 0 s,2 = fs,2 + θ = 5 + 1 = 6 f 0 1,3 = f1,3 + θ = 4 + 1 = 5 f 0 4,t = f4,t + θ = 8 + 1 = 9 f 0 1,2 = f1,2 − θ = 4 − 1 = 3 f 0 4,3 = f4,3 − θ = 1 − 1 = 0 ✤✁✥✁❆✁❢✴✾✁➉✁➊♠✁➋✺ ✼✁✽❈❧✁♥ ❍❇■❚✁❢✁✪✁✫❊★✁❋✁●✴, ➇ ❚ 10-5 ❯✁❱: ❚ 10–5 ⑦ ❚ 10–5 ♣❑✁➌❢✁➍✁✬✁✭✁◆✁❖, ➎✁✱✁✳✁✴✁✵✁✺⑩➏✁➐✁❞ S ✬✁✭ (0, +∞), ❣✁❤ S, ❞ v2 ✬✁✭ (+s, 1), ❣✁❤ v2, ❞ v1 ✬✁✭ (−v2, 1), ❣✁❤ v1, ❞ v3 ✬✁✭ (+v1, 1), ❣✁➑ v3, ❆ (v4, v3) ❢ f4,3 = 0, ❆ (v3,t) ❢ , f3,t = c3,t, ➒♠✁➓✁➔✬✁✭✁❸✁❹, ✬✁→✁◆✁❖✁➣✁❘▲● , ↔❘✁↕✁➙✺➛❦ ❚ 10–5 ❞✁✲ ★✁❋✁●✴✁✿♦✁➜❍❇■★✁❨✁❩✴✁✺ ❨✁❩✴✾♦ : v(f ∗ ) = fs,1 + fs,2 = f3,t + f4,t = 14. ➝ ③➞✬➟✭➟★➟✦➟✧➟➠➔ V1 = {s, v1, v2, v3}, ⑥➟✬➟✭➟★➟✦➟✧➟➠➔ V 1 = {v4,t}, ❦➟⑨ K(V1, V 1) = {(v2, v4),(v3,t)} ❃➜ ❍❇■★✁❨✁❭✁❪, ✥✁★✁➡✾ c(V1, V 1) = 14✺ ➢✄➤❋➦➥, ❨➦❭➦❪➦★➦➡✾❩➦❭➦➧➦➨➦❨➦❩✴ ★✴➦➩✁✺ ❨✁❭➦❪❃➫❍❇■♣❻✁➭❨✁➯➦➲✁★➦➳✁➵ ❆ ✺ ♦✁➸✁➺✁❨✁❩✴✁➩, ➻✁➼✁✳❩✁❪❅♣❇❆✁★✁➡✾ ✺➽➄◆✁➾, ➇✁➚❨✁❭✁❪❅♣❇❆✁★❻✁➭✁➪✁➶✁➹, ➘ ➴✁➷❨✁❩✴ ★✴✁➩➶✁➹✺