+ g, K6 L2 _- B$ x7 h* J4 `cncT = [To,Te,To,Te,To,Te,To,Te] . Y" s7 ^ M: Q' l- Ntm = [: K6 @0 v( l) _8 h
[0,0,d1,d1,d2,d2,d3,d3],) x$ J# M- _6 q
[0,0,d1,d1,d2,d2,d3,d3],- n' O# w! k/ ]) Q f8 C4 f
[d1,d1,0,0,d1,d1,d2,d2],; ?) p6 v5 i) Q) L8 X
[d1,d1,0,0,d1,d1,d2,d2],, G P7 d( b/ u1 h" Q1 [
[d2,d2,d1,d1,0,0,d1,d1], ?# h" J1 _! `% n; R# ]( a8 w9 w/ ? [d2,d2,d1,d1,0,0,d1,d1],6 b2 J' d3 T+ z
[d3,d3,d2,d2,d1,d1,0,0],; @% N- S3 S& _2 v0 p
[d3,d3,d2,d2,d1,d1,0,0],0 e2 b2 ~5 }9 F7 D; }( N
]! b7 v0 j: S$ [6 h: s5 w
Type = [0,1,0,1,1,1,0,1] # CNC刀具分类 8 Z6 R* Q3 {- [- S$ G' t( f$ j; ]
A = [] # 储存第一道工序的CNC编号 , ^- N, _3 b% D ]5 MB = [] # 储存第二道工序的CNC编号 ; r- h" o' G8 x! J/ z# jfor i in range(len(Type)): % W0 F4 t1 c- u- F5 a if Type: / p! y1 J' S" Q7 J, J B.append(i)! U2 \$ Q' L) d3 h8 w+ O5 V
else:+ s# \5 [3 y0 n. Q" T6 _8 G. ?- L
A.append(i) 2 h& {$ ~9 q/ [2 ]* @" }# c1 t 4 S- k( O9 Z$ j6 U% c6 b# Sdef 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)% M, ]% Y0 e8 d/ y! ]* e j/ a0 m: U( x
state = [0 for i in range(8)] # 记录CNC状态(还剩多少秒结束,0表示空闲) 3 x' t) K6 g: z/ Q, y' ] isEmpty = [1 for i in range(8)] # CNC是否为空 : q- E8 y Z2 G log = [0 for i in range(8)] # 记录每台CNC正在加工第几件物料: |& B4 x. S0 ~& q' `
count1 = 0 % D% Z2 _4 j {0 \ rgv = 0 # rgv状态(0表示空车,1表示载着半成品)2 W/ Z* f j; W9 a( V
currP = 0 + [, M/ g$ Q3 q0 Q- [* n% H total = 02 t# e5 P. N K( X6 _
seq = []8 V. l# ]0 F! a# @: ~
flag = False ; \1 T/ F F- g5 B for i in range(len(Type)): - a; {# W9 T1 q; V, O$ d) D if Type==0:9 J' Y0 f5 l4 x, P- C5 v
seq.append(i) & [% [9 B! m5 q3 F& L' s flag = True# V! ], L7 y8 W; A! j& w
currP = seq[0]( Y$ ?8 S$ q% p0 H" v
seq.append(currP) @. w$ W: D1 C/ _9 U
count1,rgv,currP,total = simulate(seq,state,isEmpty,log,count1,rgv,currP,total)% I& Z# a: D8 m5 N
return state,isEmpty,log,count1,rgv,currP,total,seq f$ m" n9 X$ F* F $ W6 F- B& _4 y8 V# ~( ?& H \4 q" f) p( idef update(state,t): * G7 `% Q& W- r for i in range(len(state)): $ F# X; V8 S) ]# k, @# } if state < t:* r7 S+ F! `% A. D( E% {& R# T
state = 0 ! U: \ J- P& G8 { else: " p) G8 X. z5 H6 T- K% e state -= t8 l1 X/ g/ Y4 t& O# d& W+ t! u
/ {9 T9 h) _: Y4 {3 _% }# Ndef 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) ; R+ K- \ ^5 `) s index = 0 , M |5 } B! R, a$ e temp = 0 , N/ Q+ j' b3 R/ l7 ? pro1 = {} # 第一道工序的上下料开始时间. R/ }3 I% k5 z6 G8 G
pro2 = {} # 第二道工序的上下料开始时间- k& A# G4 C9 G- t# d' C
f = open(fpath,"a")4 D* v* ?# F0 m5 A2 ^
while index<len(seq): . U+ K( R* x/ d0 r0 G print(isEmpty): W) s$ k3 @ K: ?
nextP = seq[index] " t8 s3 Z- p0 o8 O0 r' h t = tm[currP][nextP], s; V2 |5 \6 o6 u. F7 C0 t
total += t 6 {( A( j& J: `) I. r update(state,t) 7 Y9 v' y& \4 E& h/ B; L if Type[nextP]==0: # 如果下一个位置是第一道工作点 : G D, g$ y* v# ?, z2 {3 V* I- a count1 += 1 3 @6 [3 ]7 Z9 _2 ]5 }* H; n6 ^2 t if isEmpty[nextP]: # 如果下一个位置是空的 : u# H$ X R: K% D$ u 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)) $ Q5 R' H* @4 [$ ^ t = cncT[nextP]1 E: f4 K5 l u, V; e. f
total += t S! f1 q$ n- x: C Y F update(state,t)' F& C1 }$ y9 R; H- K
state[nextP] = T1 # 更新当前的CNC状态3 S2 N1 N) R9 H+ D4 D& ^% D
isEmpty[nextP] = 0 # 就不空闲了7 k* S C x( I8 ?# ^% o( Q
else: # 如果没有空闲 7 Z1 O6 w, g, }( {+ r! E; V0 { if state[nextP] > 0: # 如果还在工作就等待结束! c4 _2 C q" P& v, v' C) V
t = state[nextP]( i- q% s2 Z" p6 J( k3 ~8 \
total += t6 Y; J) d3 v9 }4 U- m4 V* W5 _1 v
update(state,t): |7 j' D- U5 G. ?' d/ c" [, y) E
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)) / s8 {, S1 d5 Z& O 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)) - _, g) W6 e& e E* R& m# ^ t = cncT[nextP] # 完成一次上下料 ) @7 E% p: n a' b5 S, q: a4 g total += t % I p6 z& S) z& t V update(state,t) + C- r, u5 P- _' A state[nextP] = T1 6 K% c+ n1 l& v8 ~, z4 R. O' a rgv = log[nextP]2 {7 E$ l Q: w3 h% J) h$ _
log[nextP] = count1 ~% [ r8 m. w7 ^1 y else: # 如果下一个位置是第二道工作点 - W) O& @% ~. r" `' j1 f" b if isEmpty[nextP]: # 如果下一个位置是空的 }2 f/ n. t6 [2 [. a 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)) 9 h6 X) ~1 [7 |5 X1 C1 x" J t = cncT[nextP] ( n" @9 t* `6 K+ p } total += t 3 v4 P% o# e; ] update(state,t) & N4 \! G0 h. }0 B& _ state[nextP] = T2 " f2 D! N5 x" h isEmpty[nextP] = 0 ) u" \7 A0 d& ~: `6 A
else: # 如果没有空闲( }6 H; v3 N+ a
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))5 g% Q/ l* L/ d. ]- Z* X
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)) ! p/ Q, w5 Q( G* w5 Y if state[nextP] > 0: # 如果还在工作就等待结束 - O8 I" Y! ]( @4 a( Q t = state[nextP]- \( O4 H& |% B4 Z$ E7 Z& v
total += t % H7 A0 T+ n; c( X update(state,t)6 R6 {- Y2 V% _5 D& z- m
t = cncT[nextP]+Tc 5 P6 T; W7 [: O! O, a& V total += t + |5 i4 K: D) A" i update(state,t)" j4 ^9 v0 a6 P/ A2 {! L8 @
state[nextP] = T2& b# C# m1 ]2 b8 T1 d
log[nextP] = rgv1 N6 v$ F; P3 @" ^* {0 _
rgv = 0 L1 r$ i& N' y) D* S8 n5 `2 R
currP = nextP & E8 ?5 x8 c' l z0 U temp = total 5 q [, r: o/ q/ B, {) p index += 1 % {+ u4 e3 I F! M
f.close() - n! K4 E( z- W+ J total += tm[currP][Type.index(0)] # 最后归到起始点( Y d; b$ y( b4 D
return count1,rgv,currP,total& t0 b2 J" C7 \; c5 s
' P$ m5 K" G4 S0 g/ O w4 ldef time_calc(seq,state,isEmpty,rgv,currP,total): # 主要用于记录时间 1 [6 }+ O1 k/ v' V" B4 P, H$ r index = 08 G) M+ B o* ~0 C' f& x" v- D! C
temp = 0 $ h% S2 `% W" L( U7 o* C: N1 p while index<len(seq):4 ?4 ?) _# j4 z4 T
nextP = seq[index] 2 ~/ i0 y+ T- b; Z1 n+ M- s t = tm[currP][nextP]: R' e+ p M& V, ^, \
total += t5 o2 J& n* Z; Q2 ]; H& D( s
update(state,t)/ v. }- U2 Z& d; n3 `4 M- O
if Type[nextP]==0: # 如果下一个位置是第一道工作点 ' O/ U4 s7 @; L5 J) m8 l if rgv==1: # 然而载着半成品. R; q; i& y# U$ f
seq.pop(index) # 去掉这个元素并中止当次循环进入下一个循环 : m! K8 C! f& } continue , T/ i$ ^' a; P# G' R; F, I# V if isEmpty[nextP]: # 如果下一个位置是空的1 T$ e* t! X' u
t = cncT[nextP] + Z3 Q* T1 M6 v( n, w) e total += t/ ?: _% w. h& g% Y. N
update(state,t)* I ^; I: c3 l: r9 b: z
state[nextP] = T1 # 更新当前的CNC状态! l' @5 B9 r6 I9 d7 h
isEmpty[nextP] = 0 # 就不空闲了 : j& ^3 _5 K* ~: T* \ else: # 如果没有空闲3 `. H% p3 G. m* G1 z9 P# K
if state[nextP] > 0: # 如果还在工作就等待结束' M% C. O* k v Q' m7 l
t = state[nextP]3 h# c0 Q% w0 |6 Z1 F+ V- C5 k* |
total += t4 h3 {) D8 m! {% W4 Y
update(state,t) a# A- |, w4 g6 }4 z t = cncT[nextP] # 完成一次上下料 6 J' _# {1 F# _# O total += t - C; |! Y8 _. H" q" h1 N update(state,t); ~0 ?4 m9 s1 W! g8 k
state[nextP] = T1 , C* K9 D4 @; F$ @: m5 Q4 f rgv = 1 : S, o6 Q& e" G; f0 Z else: # 如果下一个位置是第二道工作点: {" E+ i& ~6 I) c" }8 Q5 c
if rgv==0: # 如果是个空车 % \, ~5 h' S+ g( e3 h1 o seq.pop(index) # 删除当前节点 & P8 B: {2 {8 B% f continue, Z! @2 B# a8 g. O( m6 W" O
if isEmpty[nextP]: # 如果下一个位置是空的$ j7 p$ g% ]8 |; ^( I
t = cncT[nextP]# M* T, V- F7 [) n. \) w
total += t4 z7 y1 L) @" F0 _) D
update(state,t) 4 g* R! B w$ @7 z+ t# ` state[nextP] = T2 * [# a N; S& P+ J; Q6 ^5 c isEmpty[nextP] = 0 / ?7 }: J7 C; B" { else: # 如果没有空闲, Q+ V' z1 e. N8 l: W
if state[nextP] > 0: # 如果还在工作就等待结束 $ `" @7 \$ D7 ^' P8 {8 J4 V t = state[nextP] ; ]/ ~3 ]) p% ^6 u# a2 s total += t 6 J* r1 N5 k/ w; ~; R9 J; D+ Y- D update(state,t) ) u8 I9 a0 d J4 P1 R @ t = cncT[nextP]+Tc: X1 Y- @6 ^. Z7 B+ l. Q
total += t' D: Y9 e+ p' ^9 A/ {
update(state,t) ( ]8 `! W# e- @+ R state[nextP] = T2, r+ O8 W3 l; \" M2 t7 t
rgv = 0 ; o- `4 {/ F8 B* i% T currP = nextP ' |5 }) F6 H( g, J6 p& I) O temp = total ! F2 o P6 w( |2 m, k3 @6 b
index += 1 " S) Z7 K1 l5 l b return rgv,currP,total- l- t7 q( ]( x3 x
- N# ]. Z1 _: \, q8 x( [
def forward1(state,isEmpty,currP): # 一步最优. v/ W' x! ]) ^! i% {6 M" L! u
lists = [] # \0 F2 A2 p' v$ k; C if currP in A: % I3 g, T( [. W) i8 p1 \- O/ X rgv = 1 # O* e( l( a, _ for e1 in B:! L4 i6 d4 [* ?! c9 |" l
lists.append([e1]) ; Z7 _* j! ^0 {# I5 W2 Z ~ ; ], Z+ ^6 ^# U; w$ n. S3 l a) e
else:4 S/ }9 N! g/ y% y Z4 e
rgv = 0; C5 B% w( u j& Y, s' i* e% r% [: Q* X
for e1 in A:9 N% `: b0 y4 n& X1 W* y5 a
lists.append([e1])0 t( S, C; @# M9 |! f1 Z* J- b2 K
* {4 d7 }( B! ~
minV = 28800" |+ l! j% h) }3 R/ _
for i in range(len(lists)): : {1 g0 e9 F7 S* p. g$ o0 @ t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]3 y& n$ v# R K u
if t<minV:: o& d1 L9 D# Y7 ~' V# H P
minV = t2 v, W( N2 x- |$ _
index = i$ _* Q' a9 s# c8 D
return lists[index][0] & A; J8 _& L% z+ ` ' w) S% f% o0 h! O2 ^def forward4(state,isEmpty,currP): # 四步最优1 j2 R- @* u: |2 V' S& ]
lists = [] $ c& Q4 r5 F% v! t' z+ n2 U """ 遍历所有的可能性 """- t8 J* Q. u* |
if currP in A: # 如果当前在第二道工序CNC的位置 0 l- K7 O1 ^0 @; e rgv = 1 ' s/ y" E) u. L2 d! u1 U for e1 in B:) V1 h! l6 H4 \6 v8 P
for e2 in A:* x! K! t4 P% I' ^- E- [
for e3 in B: 8 y9 L% {( B( T7 B0 m# Q, K for e4 in A: 7 u% ~! p; ?* |+ ?6 E& A lists.append([e1,e2,e3,e4]) $ l+ _8 O( i1 y: ]# ` else: ( p4 T3 `0 z/ S1 y% N z rgv = 0 / g0 g9 v. Q: _9 t, ]! G for e1 in A:- t) c) L+ E2 p. ^8 M
for e2 in B: - O% N6 _: N/ Q for e3 in A: * L* I% _# N1 c% Y. B9 T for e4 in B: 4 j. }# v" K6 Y: P: g lists.append([e1,e2,e3,e4]) $ [, k. c: N6 [) U! \# T minV = 28800 2 ?7 [0 |5 s7 g' z4 t! _+ |! i5 V for i in range(len(lists)): / R! F/ N$ G5 r8 [, G% z t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]) u) v7 r2 a( n* |& g4 L
if t<minV:3 e* V0 D2 z# r8 j) y6 U) x
minV = t . |+ P/ |5 K' X$ R0 W$ [4 U index = i 1 d2 d0 t% e8 R& ? return lists[index][0] # 给定下一步的4步计算最优6 Y U) C1 k8 ^- X4 F
$ {2 g% B7 w# l) |, ?0 Sdef forward5(state,isEmpty,currP): # 五步最优2 t+ F6 P5 j" {$ P
lists = []1 p& |% I& }* Q9 Z, R
""" 遍历所有的可能性 """6 l0 T& V1 n$ r
if currP in A: # 如果当前在第二道工序CNC的位置3 {, ?2 Q& a J) }
rgv = 1 8 ?. [! N$ i! _8 c! H7 Q for e1 in B: : k. X& p$ b: x0 t0 P for e2 in A: 8 b. r; e( h5 r1 K for e3 in B:) ?* s% Q1 Q* G: q! [5 J7 s$ z8 ?
for e4 in A:# I/ T. x6 u% o O- t- d
for e5 in B: : p9 T8 ]9 [4 ^+ r: B) D5 `6 ? lists.append([e1,e2,e3,e4,e5])+ F+ i: P" u5 H. E C. y/ b
else: ( B2 G- W/ S7 w$ B rgv = 0 P7 r+ _9 V. i0 [/ U
for e1 in A:! B9 y9 T- @8 C, u" N* E
for e2 in B: F* u% b. ]& H4 |5 [8 D
for e3 in A: ) {# r4 A, ^+ c! E1 W$ H& A5 [ for e4 in B:# w1 ~0 _5 i" f6 D8 a9 W
for e5 in A:8 \& v) I, Z" D. [2 E& c- u
lists.append([e1,e2,e3,e4,e5])2 A, Z# `* b& S, P1 S S- q* ^7 E
minV = 28800 + v- Q* J3 m* Y0 F. p* W for i in range(len(lists)):* E/ p/ }+ }0 P" N/ \4 J
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 1 {. \2 D! X9 q( D5 T if t<minV: & V o$ q- T5 J$ D! w minV = t $ o& X- s: Y) {3 m. S8 t3 W2 o index = i " n& c8 U, u& V* V7 t return lists[index][0] # 给定下一步的5步计算最优, _+ G- r* ]8 h7 ^/ ]1 q" B
- t8 Y/ ? I2 c' E3 Odef forward6(state,isEmpty,currP): # 六步最优 ' S% h. f9 ?9 ~9 W lists = [] / U* O1 f+ Q$ L4 S" ^ """ 遍历所有的可能性 """( W+ [# p! }. q, b
if currP in A: # 如果当前在第二道工序CNC的位置 5 g7 A. Q1 u6 q+ m% U* f rgv = 1$ V5 R- S( |$ w1 Y$ X! ^1 L8 D
for e1 in B:4 t" d4 O6 `7 J' {3 Y* T" ?$ `1 s
for e2 in A: * E/ N' ^2 q% y8 A" B: i5 z for e3 in B:- I% o' ], {6 j9 t
for e4 in A:- L. B9 P' s6 c3 W
for e5 in B: * o& I9 @( n K+ T& f7 [ for e6 in A:( |: H/ p: E1 {5 [+ w
lists.append([e1,e2,e3,e4,e5,e6])& g n8 H+ @8 ], ?
else: ( Z' p9 A& C2 A9 r( ~ rgv = 0 ( k; U0 ?. n% z n for e1 in A: . F. \1 x( b) f% n' O7 u6 O7 `2 H4 D for e2 in B: ) m5 R* Q _1 U' Z9 b8 _ for e3 in A:1 H0 w' b; C: l- R
for e4 in B:8 D% y6 ~9 N* @2 \ x
for e5 in A:2 H& I; E c( {0 X+ S7 w# v& n5 T
for e6 in B:, F+ H( H8 g( \" _& m5 s
lists.append([e1,e2,e3,e4,e5,e6])! }! @4 d5 X4 i5 U4 P P* w7 n
minV = 28800# l0 R9 h- M% f0 Z' u! z
for i in range(len(lists)):$ l! B# r1 ~$ _4 p
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 9 v7 g+ c8 r' e' {, V/ x. ? if t<minV:" v& H& u( g5 ~* L# v* g1 y1 r
minV = t - t; r8 J" T" b1 X3 h2 D- }" } index = i9 T) ?% x1 d* _& ~ r, J
return lists[index][0] # 给定下一步的6步计算最优 & v7 P4 d, y9 ?; G/ x2 w, v. g( z. _& H P- i6 k0 P( r2 o
def forward7(state,isEmpty,currP): # 七步最优 $ b, h5 o y8 @3 Q: g' ?' ` l+ [ lists = []* ~8 X- l, |& P3 M3 |( o
""" 遍历所有的可能性 """4 q' A1 L: @* Z
if currP in A: # 如果当前在第二道工序CNC的位置 6 v5 M" ~9 A" z9 y, ~. I rgv = 1 3 A9 D5 c# @3 k3 Q/ e( U for e1 in B: . i! c! D4 p& H$ o8 H; r) O2 B7 M for e2 in A: " y# D/ f3 P2 F! k/ g" D M3 j) `5 d+ b for e3 in B:6 a, } @5 T; ?+ Z' O; n
for e4 in A: 3 ]7 R4 c8 B' X+ c6 ?* W9 t for e5 in B:( G, L- l* m1 a- l) J& S' ^
for e6 in A:+ f& _) {3 h/ K {# w
for e7 in B:& f [3 A' x* @7 V- f' b- P
lists.append([e1,e2,e3,e4,e5,e6,e7]) u; ~/ U/ U) [! t% q4 M% j& A3 ^
else:- h7 g- u. ^7 B5 f' P% P* H- b
rgv = 0 ) @3 x% c. p1 F4 K- a for e1 in A: W9 Y/ s$ R& y0 B3 f: ]9 A/ x for e2 in B:% ?7 g7 B# L, b
for e3 in A: 1 z0 [( i" u0 t B# U/ |- H- p for e4 in B:( a' L8 K9 v$ x3 B4 o1 t
for e5 in A:9 N* J$ Z0 r* u! c7 {! f k. v
for e6 in B: ' y" s- j" y* {7 ~5 p: ] for e7 in A: , w! P4 q& Z: @' Y lists.append([e1,e2,e3,e4,e5,e6,e7]) * f$ b- G4 t/ x minV = 288003 R# y, a5 @- h4 e: Y9 ?( O
for i in range(len(lists)): E0 Z9 `( N7 R6 u1 v# j& n1 T t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 8 X2 y4 M1 k9 _% \' X: s if t<minV:( z! f$ ^7 | g% T& E$ J- ]7 C
minV = t& e& D' e0 V) Q$ x! ^ J
index = i5 U) E$ ^ s0 `. H. J: i- k
return lists[index][0] # 给定下一步的7步计算最优 5 n& J) N: L# o9 a! H # B1 p/ e) n3 ?( W8 L4 Mdef forward8(state,isEmpty,currP): # 八步最优 4 D. K) `! S* s8 T1 U9 c lists = [] & q4 c) s2 T$ F """ 遍历所有的可能性 """. i, s/ M7 _; n" n. |7 n
if currP in A: # 如果当前在第二道工序CNC的位置 ! @: ~5 v1 ?) v rgv = 1 9 ~- ^( {4 r$ ^9 |- I$ v' }( V$ N for e1 in B: 0 R& C% y! W% r5 d! Y for e2 in A:5 E L$ B0 G" s
for e3 in B:! ]# E$ a) S; a
for e4 in A: ! E1 X* h+ j" ~0 x# f/ `9 t for e5 in B:- D! m' m! B) M/ ~: P: \" z
for e6 in A: , z$ {8 F1 [0 c% o: r$ {& t for e7 in B: 5 ], Z7 W' V- G" X$ R for e8 in A:: `, o$ i1 J, }1 i1 m
lists.append([e1,e2,e3,e4,e5,e6,e7,e8])( N0 u9 i& p$ K- U0 E
else:; }; P! x/ Y7 o" @3 M9 l" X
rgv = 0 k: i, B+ t/ @3 ]$ d
for e1 in A: 6 Q5 Y7 T- b z' X9 r N) U for e2 in B:3 ^5 a/ m0 o$ t' c+ s# `
for e3 in A:1 f, p5 D/ S$ x% y
for e4 in B:0 y \ D1 } F9 c0 R$ d
for e5 in A:' C$ E! }0 U7 u' I5 V7 i. F! T) m! N
for e6 in B:' R- p- r1 N! h7 k* X" o- G
for e7 in A: / I! {. i6 ]0 h. L for e8 in B: * m( C0 w# R3 Z7 p' G! Q7 I4 R4 { lists.append([e1,e2,e3,e4,e5,e6,e7,e8]) % D# l3 x3 m3 K5 b f7 o6 U minV = 28800$ V9 d8 Z1 {. Q' e, I
for i in range(len(lists)):/ h0 f0 M4 z" J
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]& ]) K# G' W! t1 E* X" L( x
if t<minV:3 G( w0 C9 o) @! F& c- U7 k) b
minV = t$ k) [* E2 R8 p, U6 S2 |6 t
index = i7 y( p& b5 M. x& o: G3 _# {
return lists[index][0] # 给定下一步的8步计算最优( {& W; S- d; w( v! c7 |
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def greedy(state,isEmpty,rgv,currP,total): # 贪婪算法6 B( L5 p/ x: w/ e
line = []9 C/ V {7 K7 P" u; Q- J; ? V
count = 0 + e5 c0 r- R! C* f, [; n4 N while True: 1 |: W: I% w$ @' W) o% g. d5 F #nextP = forward4(state[:],isEmpty[:],currP) j0 c# ]9 S" ]' @ nextP = forward5(state[:],isEmpty[:],currP) * ~1 t( e2 |- o; V0 n
line.append(nextP) " _! I! [. z$ C4 }) j" @ rgv,currP,t = time_calc([nextP],state,isEmpty,rgv,currP,0) 9 _" ~2 |. a) q- \5 f5 J% ?: A total += t& z) j$ H" k3 n% T3 L
count += 1 : R/ Z- _# s9 d5 a, l if total>=28800:* Q) _9 o t% R8 V: Y R
break 6 c& P$ W7 b- L9 a/ N8 Y" I8 l return line1 {% |5 g j, R+ Y% k+ n. E0 i
" L3 H+ u% w: D9 }% k$ S+ m
if __name__ == "__main__": : g0 J9 `. ~- R state,isEmpty,log,count1,rgv,currP,total,seq = init_first_round()0 V- o+ B A4 L4 x
print(state,isEmpty,log,count1,rgv,currP,total,seq) 8 `( A- N5 V2 [8 n5 v line = greedy(state[:],isEmpty[:],rgv,currP,total)9 p( k4 i" c# v7 M ?
simulate(line,state,isEmpty,log,count1,rgv,currP,total)) X2 G/ G _* e: U. O
% K& V7 L( T( q) [8 g
write_xlsx()( ]9 |* M" q I v: A N
后记- S2 n$ y G" x. ^$ U4 O- C+ A
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这次博客有点赶,所以质量有点差,很多点没有具体说清楚。主要最近事情比较多。本来也没想写这篇博客,但是觉得人还是要善始善终,虽然没有人来阅读,但是学习的路上还是要多做小结,另外也是万一有需要的朋友也可以给一些参考。虽然我的水平很差劲,但是我希望能够通过交流学习提高更多人包括我自己的水平。不喜勿喷!$ e& {, x" ~7 h6 i& r$ p
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