2 H$ Y4 B2 c1 ?, @""" 5 ]# W1 M; n4 i2 _: v5 U; g8 rd1 = 18 4 T( o' @7 P' }. o' W* qd2 = 328 F8 [' }; e5 A$ x0 J8 K0 s
d3 = 46 3 N5 K: H7 i* Z1 \# DT1 = 455. n. {: H4 } {1 \5 a; O
T2 = 182+ R( S- \2 ~& z, e0 p0 K/ N5 y
To = 27 % z- G N) t( j" y5 B0 VTe = 32. n6 l9 m8 [6 l
Tc = 25 . o" h( b5 e" J% c""": } F! G2 v( f
Y& \$ o) _0 L, u& O5 ^, EcncT = [To,Te,To,Te,To,Te,To,Te] 5 t- y0 ~# ^% J# t* O3 i( K Utm = [: Y& J% [. h% C( I! g" Z1 M0 b
[0,0,d1,d1,d2,d2,d3,d3], + ?8 }* \8 M- S2 m0 }2 U2 K [0,0,d1,d1,d2,d2,d3,d3], ) i V. t/ P" D, k [d1,d1,0,0,d1,d1,d2,d2],4 Q" t5 P8 I4 w' a" ]
[d1,d1,0,0,d1,d1,d2,d2],( Y( R, ?) h& h% o# `
[d2,d2,d1,d1,0,0,d1,d1],7 c3 s( k& |+ Y' ~" U* p, m
[d2,d2,d1,d1,0,0,d1,d1],: T8 r4 r% C' n* Y& w# Z/ T
[d3,d3,d2,d2,d1,d1,0,0],- ~% a9 _1 K. z4 Q! A: S- ? v# x
[d3,d3,d2,d2,d1,d1,0,0], 8 o% b( j! x [. S, H]( i- W x; [4 w3 H2 g; C
Type = [0,1,0,1,1,1,0,1] # CNC刀具分类: j7 o3 H0 {0 U$ }. w
& s3 K5 J$ S7 ]2 A
A = [] # 储存第一道工序的CNC编号 ( e* b2 x8 [; N3 T; t6 ~B = [] # 储存第二道工序的CNC编号 ' Q' s: S v+ y3 r$ _/ Y Nfor i in range(len(Type)):* e$ E0 e0 F. e/ N
if Type: . n- T$ J5 x0 r* s" |( u" C7 `* m r- g B.append(i) 1 r8 M6 G7 G/ }. \ O: A else:. S/ B4 }) W% S3 c6 z% Z* N E
A.append(i) . L3 F0 D8 l" @" s& j5 o e/ h+ v& L9 r2 J5 w9 a: t
def init_first_round(): # 第一圈初始化（默认把所有第一道CNC按顺序加满再回到当前位置全部加满） ) A- I# _; }5 m6 g M$ L/ K state = [0 for i in range(8)] # 记录CNC状态（还剩多少秒结束，0表示空闲） - @" o5 F; G' S- ^# F/ T isEmpty = [1 for i in range(8)] # CNC是否为空 . b3 a6 q, p- w. N# A7 S log = [0 for i in range(8)] # 记录每台CNC正在加工第几件物料 1 y4 _8 A4 N" w, x; c count1 = 0 8 \- `. Z- J9 X* b rgv = 0 # rgv状态（0表示空车，1表示载着半成品） 4 C, E9 u' @- Y# p% k7 ~% _ currP = 0 1 U9 [+ h u5 ]* f) _) m total = 0 0 c$ _! u$ U6 k5 l. s e seq = []& f t; \. t& N! u6 O" h6 k5 N
flag = False 9 r' P, A; E) T2 s- g for i in range(len(Type)):! J4 a% C& ^4 I7 L
if Type==0: 8 l, Q; g' W6 d' q1 o5 v. M seq.append(i) 0 I6 a, p% V3 B5 F2 |: y flag = True 4 e; ~+ h' \3 i9 e' H# B2 O/ Z currP = seq[0]5 Y/ z0 i3 E% m# ?) X4 _) U8 r
seq.append(currP) * a; J. Z+ r+ M count1,rgv,currP,total = simulate(seq,state,isEmpty,log,count1,rgv,currP,total)- j' c# d3 K* R% N2 K
return state,isEmpty,log,count1,rgv,currP,total,seq$ E' h; N" S9 H6 m" w- q
7 I3 x u) e& y, u# ldef update(state,t): ( _- F" p0 H2 B for i in range(len(state)): 7 u/ z9 r# O9 F( S if state < t:: c! i( A1 t" o) }6 _" x5 a
state = 02 E! G5 D" T0 b0 l5 Z) d
else:3 |8 u9 B' n' `. M4 |
state -= t 0 Z" s; F6 ?. I w * X$ m9 k6 t4 _& Tdef simulate(seq,state,isEmpty,log,count1,rgv,currP,total,fpath="log.txt"): # 给定了一个序列模拟它的过程以及返回结果（主要用于模拟并记录）% s# w4 \3 t6 _0 l
index = 0 ' n4 s( _ I# @ temp = 0 3 Q! {$ d8 L2 P pro1 = {} # 第一道工序的上下料开始时间; \! p. H# l* |4 V: R+ \( w5 @. H& v2 Y
pro2 = {} # 第二道工序的上下料开始时间# A7 K$ F" h- s2 L: y+ W
f = open(fpath,"a") $ N& d) A, s: i while index<len(seq): " D2 _7 u+ M+ ]- ]$ z print(isEmpty); h% C0 n8 b+ j8 t& Z
nextP = seq[index]7 `+ H2 j' s/ g( v5 ?: w
t = tm[currP][nextP] " P/ B7 {: T; C9 }* h. j/ n) s3 [ total += t+ \$ u- U) [ w4 Z$ R
update(state,t) 2 y& j* ]& }" c% ~ if Type[nextP]==0: # 如果下一个位置是第一道工作点# x2 g5 _5 B7 a& E
count1 += 18 N7 M1 H3 R0 q5 E/ d* [# C7 ] n" r9 E
if isEmpty[nextP]: # 如果下一个位置是空的+ H! J* K. n/ u- y6 l0 G% M# c6 r
f.write("第{}个物料的工序一上料开始时间为{}\tCNC编号为{}号\n".format(count1,total,nextP+1)) ( L, o8 U0 O4 q t = cncT[nextP] / f, l0 D9 J* d* u0 y& I) a; | total += t # m, V4 C ~% P! e; a update(state,t)" S! I& R# g8 N2 S
state[nextP] = T1 # 更新当前的CNC状态" I5 ~* t* f, I) X' G$ `6 y
isEmpty[nextP] = 0 # 就不空闲了6 F; [; @+ ~" K
else: # 如果没有空闲 O3 \$ ]2 I }( T2 h" [. Y if state[nextP] > 0: # 如果还在工作就等待结束, g- d/ F, @' |# j" Z3 K" |6 m" r3 b; ?
t = state[nextP]1 ^; \! r7 Y ^2 p
total += t7 n( j# U$ V9 U9 n" p
update(state,t)+ ]% \7 N$ f& Y7 v8 u8 k
f.write("第{}个物料的工序一下料开始时间为{}\tCNC编号为{}号\n".format(log[nextP],total,nextP+1)) , f$ R3 O( v, w0 H* r f.write("第{}个物料的工序一上料开始时间为{}\tCNC编号为{}号\n".format(count1,total,nextP+1)) " Y+ S! @ i1 }1 f& u+ k t = cncT[nextP] # 完成一次上下料+ \! a# x9 r" K6 W
total += t& V, \9 b- C9 J
update(state,t): a3 J, m | e3 ?3 C
state[nextP] = T1 ; k9 |5 k, [: f3 x$ q rgv = log[nextP]8 u3 g9 {& M& [# m8 K, w
log[nextP] = count10 B5 ? }9 I C! |' @
else: # 如果下一个位置是第二道工作点% w; u3 {$ H, ]
if isEmpty[nextP]: # 如果下一个位置是空的 7 v) ^" ^8 A& W6 n5 c: r2 b5 {0 M1 \ f.write("第{}个物料的工序二上料开始时间为{}\tCNC编号为{}号\n".format(rgv,total,nextP+1))0 [7 o q5 {) ~7 a0 V
t = cncT[nextP] ! W- c& M; z; B8 n. X' y total += t" r) Q+ M2 V' _
update(state,t) # c n, G" V7 b* i7 |* N state[nextP] = T2 & ?3 a. X3 [0 S# Q0 O0 w" d isEmpty[nextP] = 0 0 x: }' t! ~( ?+ y# J1 x: {6 T
else: # 如果没有空闲6 J4 q3 o8 u# @( S8 f p/ U/ f2 n
f.write("第{}个物料的工序二下料开始时间为{}\tCNC编号为{}号\n".format(log[nextP],total,nextP+1))' Y, E- v0 c9 D
f.write("第{}个物料的工序二上料开始时间为{}\tCNC编号为{}号\n".format(rgv,total,nextP+1)) ; G) q( p; {$ ?/ R. E/ ^ h if state[nextP] > 0: # 如果还在工作就等待结束 " B/ y7 C2 }: i. V! ^) W t = state[nextP] ) ^) V! A5 X) r$ `! M' |: _ total += t , j4 E% Z5 E5 J8 B update(state,t) % k/ g* J+ \( z6 A t = cncT[nextP]+Tc 6 f9 I& h4 u1 d Z: a total += t2 f8 n/ ~* ` w# ^4 I% d
update(state,t) ' i9 q4 g6 d& u1 c% Z3 q, D: F+ C state[nextP] = T2 ! E( `$ v: q9 S3 C, ~6 p8 h log[nextP] = rgv: _0 r: H. ~- z8 N
rgv = 0 # i, D. @7 {. K currP = nextP t4 y$ ^" ]: D4 P3 G
temp = total ) x9 E2 f6 \) m9 M# q1 J
index += 1 $ Q6 f8 ? s, w- Q f.close() $ V; R% R. r+ B" o- t total += tm[currP][Type.index(0)] # 最后归到起始点 . w2 j4 y% e0 j: b( ` return count1,rgv,currP,total4 H1 X2 Q2 M; ^& q2 a! O
9 l/ L) P( _& r2 t+ idef time_calc(seq,state,isEmpty,rgv,currP,total): # 主要用于记录时间' V0 k' e) ~0 O- h# E
index = 0 1 v- i% Z. B3 j temp = 0 5 Q1 ~' w' T) J7 J* o while index<len(seq):! e( a' @$ ]* S! G# A$ f( k
nextP = seq[index] ) B- Z7 C- X6 S& A1 R6 z t = tm[currP][nextP] % {; Z4 D% P5 i! | j% @+ M total += t. t/ @# l3 R: P1 Z% K5 v& b# U
update(state,t) 3 ^6 B4 g; N c# w/ ^- ~" c, y if Type[nextP]==0: # 如果下一个位置是第一道工作点 - ^0 J" M9 M( V# a if rgv==1: # 然而载着半成品 ( ?& M5 X9 e, Y$ V3 [) h) A( } seq.pop(index) # 去掉这个元素并中止当次循环进入下一个循环! X2 u: J# Q" G4 J! ?9 A' _! D+ s! g! K1 {
continue 7 k3 m9 H; c( s. B" @% Y( o6 m
if isEmpty[nextP]: # 如果下一个位置是空的* T) c, r0 d! E4 \
t = cncT[nextP] $ y4 C( @# t3 i4 t J" ] total += t - f# n) ]& v7 F update(state,t) $ @! t% c; S9 ?$ S% w7 o: X$ } state[nextP] = T1 # 更新当前的CNC状态 ; D4 `7 t) n: s0 \% `' u isEmpty[nextP] = 0 # 就不空闲了 " {& ^: S& h: A4 D% |: x$ p! Y else: # 如果没有空闲% S- ?' O! O1 b8 a: c
if state[nextP] > 0: # 如果还在工作就等待结束 2 R' P0 V9 b: P; V' y t = state[nextP] + P) c$ O0 S4 Y( |) ?- D total += t- _ {# l2 W/ u
update(state,t) # ^# |' G! R" X! [8 u3 w t = cncT[nextP] # 完成一次上下料 * {6 f4 L; k- e8 d% f7 z total += t 8 t" P2 S- x) `, j/ ~, F update(state,t)1 M$ u9 T7 M: e$ Y- v) z6 A2 f
state[nextP] = T13 |* u- y; s& ?' i* C, R) x! D9 v8 a
rgv = 1* q0 i ^: I# s& ?, `
else: # 如果下一个位置是第二道工作点# y8 L6 t4 T5 H! e" x; X6 V
if rgv==0: # 如果是个空车! [ T- [8 W+ d! V; v: D
seq.pop(index) # 删除当前节点 7 _9 n2 j* d& d4 g8 r' o5 Y3 h continue 8 ?: o$ H9 w# |9 K& S6 o% ] if isEmpty[nextP]: # 如果下一个位置是空的8 i6 }( \9 S7 `
t = cncT[nextP] 2 L+ R# C: Z9 z$ d5 i, s3 X2 C( j, | total += t ; M0 ?; \4 c0 _, u8 [ update(state,t) ' s a( g: m" D, M7 A8 n. ?& X; j state[nextP] = T2( I' F- \: j" ]/ Q0 ?" H
isEmpty[nextP] = 0 5 ~) q# x, }" N! S$ C
else: # 如果没有空闲 % C P2 u7 U0 ~0 H) i0 X4 f if state[nextP] > 0: # 如果还在工作就等待结束0 ]& D" \4 W8 o3 X& D) }! B7 _
t = state[nextP]% W: [# z3 g* {% |) L: X3 Z G
total += t / k, [7 [) Q. l5 m) b update(state,t) * X5 j. F" r- c1 g/ I7 `! X3 [ t = cncT[nextP]+Tc: o9 Z( G7 P' z% K6 Z9 y
total += t6 a* m1 p4 L8 v, p/ P+ g% z
update(state,t) % Z- T& M% c3 s+ v0 w/ i) w X state[nextP] = T2 - A( A D6 X5 O" B/ [ R9 G+ I rgv = 0 * j Q& e' R0 E! p6 d$ w currP = nextP" B- m/ i: w: X
temp = total . \8 r# W) q6 c) @) P. Z index += 1 ) }: b0 ]! H+ ^1 V# ^* j return rgv,currP,total 3 t$ z( Q1 }1 p4 l- Q" P: E1 x ; o; l6 u, _2 K/ c5 [def forward1(state,isEmpty,currP): # 一步最优 & i/ {% F/ F6 T! m% }0 I: | lists = [] 5 [5 N9 w7 l4 P+ Y. y! c2 y! o. w3 x0 F2 j if currP in A:) F+ g+ z9 F) Y7 H8 A
rgv = 1( ]! L) C% M6 [ d
for e1 in B: ( n. X4 t# H) @& ^: W, v lists.append([e1]) + t7 ]; L* H; U. x + s& l, O3 W2 x" G6 W( f: x/ i7 g4 q
else: 2 X+ u$ F" j9 H) m rgv = 0 0 n+ g) z& F# d' v for e1 in A:* S: f/ [7 t3 m, U3 L
lists.append([e1])1 g6 j) J! \9 D0 q* P% K
; Y0 K. o1 P2 J1 }( r minV = 28800 , B) c9 x9 Y j$ d. L# V3 C! s, t for i in range(len(lists)):7 Q; |# _8 A9 J9 Q. f1 v5 o; y* i
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] " E' M' H$ f; k$ D! K; X( K if t<minV: ; l, X- ~7 m7 E) R- A9 Y, ? minV = t 4 }& F" T0 o. b* ^4 ^ index = i 8 N2 O) k+ b1 q9 S4 b return lists[index][0] + s0 P+ S' c0 f; I4 @3 j( b2 X0 _; ?- N$ U
def forward4(state,isEmpty,currP): # 四步最优2 C- h3 M4 r7 l8 `/ {! Z+ s
lists = [] 8 A' x/ Z( M/ r' n6 Q """ 遍历所有的可能性 """0 M2 m: R+ [, O) i
if currP in A: # 如果当前在第二道工序CNC的位置6 i+ F& U) h: M$ o6 B
rgv = 1 ) K0 R2 r; `- { for e1 in B: 1 B6 R$ q" ]; ]% g. u- I! T for e2 in A:; _8 I8 L+ r6 Z" J0 z1 b7 @2 R
for e3 in B: / N$ z& i5 a! Y* j8 F0 i+ @4 G2 [ for e4 in A: k0 [9 g( L5 E lists.append([e1,e2,e3,e4]) 7 Q: M' C+ r' M! f, w else:; P0 C1 n( x& r. F) h
rgv = 0$ B0 |4 }- y1 Y/ ]8 U
for e1 in A:& d1 n: G- j+ q! k. P
for e2 in B: # x5 \# D" b& F9 ]9 F for e3 in A: - o5 T$ M1 |2 N% _ for e4 in B:. r& D: S8 N0 N! l7 @$ M; n5 i
lists.append([e1,e2,e3,e4]) , e( y9 C3 X0 B minV = 28800 - o! {! B# i. Y: ~2 h0 q2 U for i in range(len(lists)):5 R5 V. o) E; U: E
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] & l* p1 S+ N8 B# t4 U if t<minV: 0 q; r) ~$ I; B% D/ [" t: K1 U+ ` minV = t) [6 F q8 U! h- k# X) }
index = i 0 z0 F; J( R+ i1 w3 ^ return lists[index][0] # 给定下一步的4步计算最优 8 Z8 \" t1 c/ a: w 3 V# o3 r. P) L8 t9 Sdef forward5(state,isEmpty,currP): # 五步最优 # j* S! V7 \, {( `; G lists = []9 X W8 m/ D1 L! w" h7 i6 L
""" 遍历所有的可能性 """ 2 [; X0 P& {3 @/ H3 a if currP in A: # 如果当前在第二道工序CNC的位置 1 O* O' u/ N. y/ @$ Y0 @2 d O) c rgv = 1 # |+ j" Z- {7 q$ ] for e1 in B:/ [. E k$ H8 X/ N3 Z" ^ @
for e2 in A: 2 k& q+ S- [9 L& j for e3 in B:) N3 V+ |$ b) }! k, E# d- o A
for e4 in A:. \1 Z7 l, q( W, Y9 Q
for e5 in B:: R* x0 T$ y! {2 Z, Q
lists.append([e1,e2,e3,e4,e5]) ) r2 T0 y9 o" ~7 x3 C# _% B' W else: A' ~; s+ g" R X2 X0 N+ R/ u
rgv = 0 & Y& a/ ]" p0 \2 p for e1 in A: ! B# u2 E. R$ B( C$ O8 h/ Z) H3 r for e2 in B:' A# z. ]6 R p* z9 [+ \# \
for e3 in A: ! W% N! g$ d" N- d q4 s for e4 in B:/ w4 u2 g) r: D; ~5 i/ v
for e5 in A: 6 D' u8 |; O* I lists.append([e1,e2,e3,e4,e5])0 J* f8 ~" }, w3 f
minV = 28800 ' A k' Q) U2 W( F8 q, ?0 b for i in range(len(lists)): % |3 q9 M4 Z# p& y3 u t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]- P. G: v0 H4 w: u
if t<minV: 2 d% R1 d; D4 K' T minV = t " Z% k* y6 R' e, K, R* ~ index = i1 H' ?8 B6 I' L8 o/ V3 p) p
return lists[index][0] # 给定下一步的5步计算最优 + D$ w C$ D1 W/ m' j ! f( [- @" O) I6 n5 i2 odef forward6(state,isEmpty,currP): # 六步最优 . I& B; Q: {8 d' p+ d lists = []' c) h& \: U$ a& M8 F
""" 遍历所有的可能性 """4 A. z/ d0 b0 a7 y r1 r: ~7 \
if currP in A: # 如果当前在第二道工序CNC的位置" @$ k7 n( w8 R) y
rgv = 1 8 G; a: ~$ u' j* x; ^9 O3 d for e1 in B: 7 Z( [/ W3 L9 m8 K7 [ for e2 in A:- `( |2 l) _% T
for e3 in B:+ O: {8 z) u6 F
for e4 in A:/ C2 r6 U' Z$ q$ i3 l
for e5 in B:; U! w% T ?/ V( D/ I% @3 a
for e6 in A:! y9 k5 k" s/ l4 x# K! N/ G. l
lists.append([e1,e2,e3,e4,e5,e6]) : O) B: Y, g6 P: T3 v P0 J else: ; q: p5 N! i0 l) m% v rgv = 0 $ g% Y1 e5 w5 r) F. U! q$ U for e1 in A: 9 l1 G6 Y) C8 ~7 | for e2 in B: X4 b' n7 q! a4 n
for e3 in A:" _, `3 i$ q+ e, X6 ]) |+ E: r0 { S
for e4 in B:, z' {" |' w; S! x
for e5 in A: ( ?- ~& o- M/ d) ]* d+ _ for e6 in B: . a M* t& O" z0 p! r' j lists.append([e1,e2,e3,e4,e5,e6])+ z# O/ O4 A2 K3 J
minV = 28800, T1 w' A" \ t4 o' {
for i in range(len(lists)): 8 O. L9 b5 m4 t$ J t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]! d# w6 Z; b- i! z0 u Z5 ^
if t<minV:- j/ X& @1 N3 R q( o( K
minV = t( E6 E& T) ]' l
index = i 2 K2 ^, H8 L( _; Z, ?9 R1 M return lists[index][0] # 给定下一步的6步计算最优 |7 M6 j' B d& A6 E/ T6 x( f; d+ b% e8 M6 ?# i
def forward7(state,isEmpty,currP): # 七步最优. s& R+ z/ a x+ f- ?
lists = []( M) V* R$ F4 E9 r
""" 遍历所有的可能性 """; D- x) \1 W& O: J# ^9 d
if currP in A: # 如果当前在第二道工序CNC的位置( J- E8 U3 y. D: m: o* K
rgv = 1 " m$ H$ ~/ E4 N' L0 J) X for e1 in B: - V: g. h* ?" u for e2 in A:' B4 N5 {6 p& q5 u
for e3 in B:9 y ^% x3 [0 A9 O, k) z
for e4 in A: ! Z N4 d& [ S for e5 in B:2 D& b1 N# U4 z3 d8 F# G2 x' j8 Y
for e6 in A: 1 u: r& Z2 Z5 j for e7 in B:, o" B/ Y E5 k5 G
lists.append([e1,e2,e3,e4,e5,e6,e7])0 `( i* H# _( B7 z
else:# v4 S* Q0 K% h9 N
rgv = 0; h4 H, T+ R% g; P3 Z6 n
for e1 in A: f" i |" h/ P% B4 |/ V# m0 Z for e2 in B: 2 m6 u7 R7 X6 W- n9 ~+ K$ z7 h' l for e3 in A:% y) M( `+ a6 K' @/ K& I: O- O
for e4 in B:2 A5 r" i( T- b
for e5 in A:' T) j8 F5 B$ J" X2 }( q, }9 }2 _
for e6 in B: 3 _6 p6 C5 V4 |5 { for e7 in A:3 E! m9 h0 \' s2 @, S! _+ s" W/ U
lists.append([e1,e2,e3,e4,e5,e6,e7]) " H7 A3 @, W2 g; Z/ r1 p! n# x- c minV = 28800 * c$ m( l) q! h$ | for i in range(len(lists)): 3 A5 b( D7 I5 k: d2 }1 p t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]! ^# g2 e, O `% K1 ]2 o
if t<minV: ! y. R1 H8 s" ]4 D minV = t ( i7 v' ]0 Y8 n. Z- L M- b index = i" m- v6 }7 x7 W) a
return lists[index][0] # 给定下一步的7步计算最优 S' ?' j& f. [ S& `1 N& Q% \) u: O
def forward8(state,isEmpty,currP): # 八步最优 * Z I! k- [) \ ~; n4 b lists = [] 7 C. N: u) D; B& |( G6 P% w """ 遍历所有的可能性 """ 3 s$ |0 O8 A+ X3 R if currP in A: # 如果当前在第二道工序CNC的位置0 b% |( r% B" B1 Q
rgv = 1: F) \. Q9 E7 g) d1 l
for e1 in B:; E7 Y. \$ z4 o; n1 U, R8 s
for e2 in A: 2 v0 X# ?! J3 I. G for e3 in B: 8 S% L6 l4 _2 w* T# S2 f+ c. C" ~7 X5 S for e4 in A:' Z2 {% I5 l- K9 C* @/ V
for e5 in B: % J: H5 _- }) i }0 n for e6 in A: ( n5 d. C! B8 ` for e7 in B:& o. \" ?* I+ H3 Q$ \
for e8 in A:7 D, j4 e2 z+ p6 j* M& w
lists.append([e1,e2,e3,e4,e5,e6,e7,e8])# n# W) ~/ i/ w6 b' p( j: v
else: $ H8 X8 V/ y$ O0 |) x% ~- Y. g rgv = 0 ( l. Q% W+ n% n3 \ B for e1 in A: # y% K! Z0 m" S1 ~0 I5 ] for e2 in B: ( x5 r6 i' N& V$ T0 g+ A for e3 in A:$ a, w3 S7 q3 }5 A3 F, X3 n
for e4 in B: 8 Y+ b0 v5 {& L( v, J; Y' ~8 N for e5 in A:9 {: U" b3 n9 ~1 l7 r5 b6 D0 {9 m
for e6 in B: ' T4 d8 n/ ^ {8 U8 \) V for e7 in A: & I) l( H+ D2 ^ E7 K for e8 in B:$ P m3 [1 B7 C( o, c, l. ]
lists.append([e1,e2,e3,e4,e5,e6,e7,e8]) / v3 Q! b6 w* t( ^8 r minV = 28800" P; S! x0 f G# z+ a
for i in range(len(lists)): ) g7 S! _/ m& s5 l0 L2 M t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] : b2 V2 I" X7 O; @3 X if t<minV: 2 ^% W' w; m- o minV = t* b6 C8 d, ?- D3 k
index = i4 V8 ]2 L- A% n6 [! n
return lists[index][0] # 给定下一步的8步计算最优 7 w, Z4 I, ]" S; t5 w9 }! c- d1 x4 v0 o! b% \3 m& \$ U
def greedy(state,isEmpty,rgv,currP,total): # 贪婪算法5 d$ D. V& ?& q# R5 l0 o* W0 R
line = [] ( V( y+ W( f$ `7 b' _% e9 G6 r count = 08 `8 H6 C5 V/ M$ u
while True: : u6 k4 s3 v+ b- r/ H6 w #nextP = forward4(state[:],isEmpty[:],currP) , d$ u* z4 z3 _4 ?" _9 y
nextP = forward5(state[:],isEmpty[:],currP) ) X# _# {# K; [+ ^3 M/ B line.append(nextP)( p% v3 G) b9 M5 C/ k
rgv,currP,t = time_calc([nextP],state,isEmpty,rgv,currP,0), c1 ] b' F% H, L6 M* K
total += t8 n' W5 P% K: f7 n8 ?; `. c
count += 1 - V. k; m* \9 H. ~4 O. c9 l5 T if total>=28800:0 N8 k# ?* u5 M1 d3 W- f" H3 _
break $ u( _) Q% K6 N8 ~+ o1 m" a return line# q/ f3 V0 ]0 w* E& O
5 W& F) p; G# I' b) \! V, n( C rif __name__ == "__main__": * _ Q( g+ e' i0 A state,isEmpty,log,count1,rgv,currP,total,seq = init_first_round()8 o5 j* u% z' k& p: |1 `4 w" }* G
print(state,isEmpty,log,count1,rgv,currP,total,seq)! d/ w/ R# f* ?$ b$ }
line = greedy(state[:],isEmpty[:],rgv,currP,total) , [% U) C" _) |2 H: z8 C7 h# ? simulate(line,state,isEmpty,log,count1,rgv,currP,total)4 ]4 j; a6 D8 S% R
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发表于 2019-4-12 16:18
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