+ v1 K. f$ t: N% a7 n问题一共有四种不同的情况:一道工序无故障,一道工序有故障,两道工序无故障,两道工序有故障。7 {5 ^& }+ B2 f' |0 [
0 x1 e0 R; k5 D. k2 [一道工序无故障 7 w! C+ N. o, L4 M, m v1 r- `& f/ [+ t$ C p! n
第一种情况是最简单的,直观上直接不停地1234567812345678……按顺序上料差不多就是最优了。但严谨地来说,虽然题目中给的三组数据确实都是用这种最幼稚的策略能够达到最优,但是如果对于一般的情况而言,比如最极端的情况下,RGV移动时间无穷大,那RGV显然就只会不停地在121212121212……这样原地上下料了。 % G/ [( |4 Y; G" }4 o. c4 F* p% {! Z) b
然而我们发现无论参数怎么变化,最终RGV给CNC上下料的过程始终是一个周期性过程。当然这个似乎很“显然”的事实却是相当难以通过数学严格证明的(参数已知的情况下一般比较容易证明,但是所有的参数都是未知的情况下是很难严格说明的)。我赛后也仔细的思考过,但是也没有得出很漂亮的证明。我最终仅仅是针对给定的三组数据使用了遗传算法对RGV前17次上下料(17次是考虑从初始状态开始循环两圈的最短路径)的最优路径进行了搜索,并且利用穷举证明了这是前17步最优的上下料次序。之后基本上就是不断地循环。& [' |9 n6 u& H+ z$ A0 s
. Z5 p; W6 R. z" v这里的模拟退火遗传算法比较鸡肋,所以我不详细说明,在第三种情况我会详细说明模拟退火遗传算法的原理。 5 n. G! `7 [( Z. v i- a# d7 q * V- e: N8 P' n" a# T以下给出第一种情况的模拟退火遗传算法算法以及对应的穷举最优证明 ↓↓↓# `4 u9 y& ]6 l2 |
# -*- coding:UTF-8 -*- R. z+ z2 C4 @7 F2 o
"""0 s) C9 H1 P6 j8 J
作者:囚生CY & A, j- ^) y6 t2 n, F 平台:CSDN: k/ y P$ \: j$ N/ L3 ^: _
时间:2018/10/09 # U- v. m, n: b W2 K 转载请注明原作者 * t# S0 p, m3 O 创作不易,仅供分享 : V; L4 X1 e( M- `, g6 G( ?" z""", t) N6 h+ x7 y# A( M' p/ R3 q
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import math 6 M0 m4 D5 E, o$ aimport random / O1 U* d6 R9 m% [import itertools: u1 s0 t: h; ]' s7 K
0 ]/ A! q! B; m( {, q9 D""" 选取一组数据 """ ! ~: [2 G K! N* f$ lT = 580 / m( D+ \/ i" G/ q$ A) @) td1 = 231 i1 g0 E5 I5 H% E$ t- ^5 Y
d2 = 41, V. g1 P! R; n, d
d3 = 592 Q" K+ Z- X; [9 i# G
Te = 359 B8 z! I/ G$ A; j
To = 30) I6 K3 s s- F1 M& m
Tc = 30$ ?) I& s3 {+ f
) q3 `; [. Y& F
CNCT = [To,Te,To,Te,To,Te,To,Te] # CNC上下料时间6 t! `. k$ h3 k
6 g, `& x, H- J" e
N = 50 k. U7 \( O: n
L = 174 p6 e$ s; c1 X& H
$ i1 S+ `" ~+ `8 s6 ~& V9 f两道工序与一道工序最大的区别在于三点: 8 F5 p* k! _. n' P , ~) j R; P% f& b0 T6 g( G" n1、开始要处理CNC任务分配:分配给第一道工序几台CNC,分配给第二道工序几台CNC?具体怎么布局?) w# v2 g5 l3 R. J$ t8 c0 z: `
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2、加工过程可能仍然是一个循环,但是这个循环将可能会非常的庞大以至于不可能直观的看出来。 3 ~; \. g9 o1 x8 t/ k# f0 E2 J" ^. v! M5 r
3、两道工序的分配已经是一个严格的NP难问题了(即理论上无法在多项式时间内求得最优解)。 7 T' `, z# z* n + w3 N5 a1 D' S8 h7 ?第一点我的想法很单纯——穷举,没错,就是穷举,除了显然不合适的分配方案外,其他方案都试一遍(虽然真的很蠢,但是我真的想不出到底能怎么办了) ' l% T2 m1 _# I' n! I, E8 m2 }5 x' b $ Q- o9 ^( H' p, R, h$ I第二点因为不存在循环则使用遗传算法需要设定一个相当长的染色体长度(我们设定的染色体是RGV为各台CNC上下料的次序,如果要考虑全过程的模拟退火遗传算法,则染色体长度大约在300~400左右)。事实上我也尝试了这个方法,结果从我写完这个算法我开始跑,一直跑到比赛结束算法依旧没有收敛[捂脸]。这里给出代码仅供参考(各位朋友要是有好意见也可以提出) ↓↓↓5 W& V5 O' d B6 l. o$ V; T
- H4 G: M- A+ n0 @% \# M# -*- coding:UTF-8 -*- " ^+ H+ C* k* F""" 6 a$ x1 R6 `1 j/ x# p 作者:囚生CY " s8 Y3 `+ x, B N. N/ g 平台:CSDN 4 M% W9 J* ?3 c/ D* g 时间:2018/10/09$ n3 L- C! S) S" y) G6 C3 `7 l
转载请注明原作者! z, L! d+ N: m: a' E3 Z
创作不易,仅供分享 4 w7 z+ m7 O, g! D"""' G, t" W% a# P5 M) t' p- e3 d! H
import random. i3 {/ Y6 ~* } W3 u! G- o
: w- s& K% g7 T5 K- v9 D5 S/ o9 [
# 第1组 ( a4 `/ ^- ]8 V"""" w8 A% s( ^+ h
d1 = 20 1 q. ~- H; F9 T& E1 Qd2 = 334 C8 T1 X# _! f, c' L
d3 = 46 3 r5 [( ~2 W9 x3 G) t$ i' ZT1 = 4007 L8 t* \+ z( ~* t
T2 = 378 6 x, _# s: W0 \; D1 G4 {6 |; ~To = 28 # {4 E- g5 F# V4 N) r- P6 HTe = 313 o2 q0 e( }- z7 @# Z* r
Tc = 25 * `* w5 W" a1 k9 g/ r9 G G7 X# v""" 3 m* t+ v2 {2 _; n' h 6 v* W8 p6 _6 A# 第2组1 {) z6 `$ V5 N. R# l
"""+ ]" u5 j3 n: r: y: a7 D
d1 = 23 1 r, K) ]: w' L2 P6 Hd2 = 41 ; h- Z+ q( X# ~* bd3 = 59! k3 q- G6 j2 z W
T1 = 280* e# u0 e3 j- q- Y) g8 e& b( u
T2 = 500- p7 v% V' x2 i+ Q
To = 30# [4 K7 M$ `5 `5 e% H+ t5 w: f
Te = 35 " K2 A$ r0 [& |$ h' iTc = 30$ t$ r7 O. F: t
""" ( E9 Y4 n# Z3 O. v5 a 2 z& C0 Z' }" w: x. g: e# 第3组& m( ?2 P9 Y- W0 u6 w* `
d1 = 18 0 V. h: H, f) I' g- ^0 B' W5 Qd2 = 32 M: A0 y# l- V5 H; T) S- Ud3 = 469 L1 \, T6 Y! l0 q
T1 = 455" N4 F; p: A* {, Y' r
T2 = 182 ! y1 c; ?& I' P% p6 }' n( aTo = 277 h9 _+ v* }: r5 T0 w% o0 T
Te = 32 ; u% I! x& y d. iTc = 25 / ]+ n: k$ T# T4 N* C# ?/ S, P1 R; | T
cncT = [To,Te,To,Te,To,Te,To,Te]: k( b# S4 l w6 Q$ ]
tm = [ 3 a, N. b+ M: Q9 H [0,0,d1,d1,d2,d2,d3,d3], + L' E# ?+ Q4 Q$ I0 N7 C8 M2 \ [0,0,d1,d1,d2,d2,d3,d3], ' c3 w5 G" E$ n- e' X: ] [d1,d1,0,0,d1,d1,d2,d2], 1 b6 O) D5 _( W0 J; D [d1,d1,0,0,d1,d1,d2,d2], ; U. v, }7 m' X- f) S: E [d2,d2,d1,d1,0,0,d1,d1],7 q4 I4 e/ q+ [" R6 [
[d2,d2,d1,d1,0,0,d1,d1],: z4 M0 u2 m+ E( H2 e S! H& X0 p
[d3,d3,d2,d2,d1,d1,0,0],8 T: v2 |; \+ I9 b8 F+ q) L" J/ w/ b
[d3,d3,d2,d2,d1,d1,0,0]," b3 a* a+ ? d& l6 I1 B
]2 i3 Q; h& o) {% L+ A' j: U
Type = [1,0,1,0,1,0,0,0] # CNC刀具分类) @' i- W8 |. g4 \ {
1 i1 v5 \/ d$ w4 I4 y
N = 646 r# ^8 T) ]; J6 n9 h# o
L = 100 $ `9 a9 _! D( c% kvarP = 0.1+ |2 m0 ]' H" Z/ e
croP = 0.6: _( I% M+ O/ b3 w
croL = 2 $ Y, j+ k' f8 X% y4 Ge = 0.99. L, l5 n0 \) x' ]
! E( c: v! s% m- Udef 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) 8 c7 C2 w. S( r% s& 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) & C$ }& d, E) z1 F, Y3 ^ isEmpty = [1 for i in range(8)] # CNC是否为空5 }; _3 m: b/ F/ z! O
rgv = 0 # rgv状态(0表示空车,1表示载着半成品) 3 a j# \) }+ B! v currP = 0 , z' e( N2 h8 [- g total = 0 ( M) C* X3 ?" q3 b' S seq = [] 9 z. H' E7 V5 L. d1 A$ F flag = False / ]1 Z* x, t1 T8 W. z, | for i in range(len(Type)): . R% ~# Z& t7 ^/ c. a if Type==0:( ~# W/ M, a4 d8 @3 {
seq.append(i), v# h" m j+ U9 D2 h: y& ?2 b
flag = True 2 Y/ R8 z! [1 e$ t currP = seq[0] 0 r( P, R3 j* m8 B# h seq.append(currP) 8 T# a! ]1 {8 R: w$ E+ l2 g" F, p rgv,currP,total = time_calc(seq,state,isEmpty,rgv,currP,total)4 h" q5 h/ E! c' a8 }: J% X) c9 j
return state,isEmpty,rgv,currP,total,seq ; p2 l% m( J/ T+ \% L- p: B 7 W% B; C& @. C: Udef update(state,t): ! H/ U" K* Y8 B; s8 E for i in range(len(state)):3 t6 j& N" }6 @; w9 _! P: i# k
if state < t:- f* y3 V* l% z' G1 j- A+ m
state = 01 M) A8 ~& Q; A% _8 d2 g7 z
else: ) D2 H2 w( n, U+ t( `/ M- l state -= t , d. u8 }, f+ S; e" M- Z * O0 `" i# I% odef time_calc(seq,state,isEmpty,rgv,currP,total): # 事实上sequence可能是无效的,所以可能需要4 ~1 c7 Y t* F L
index = 0/ ?0 S; q9 R6 L4 O/ ^
temp = 02 |0 Y7 R7 x: E5 V: }) J
while index<len(seq):& V A( f9 p2 E6 r' \+ a7 U) Z
""" 先移动到下一个位置 """. _, m7 |9 m; W) P5 @
nextP = seq[index] - J( f9 ~4 Y: n( I; K t = tm[currP][nextP] : R' w8 Z6 l$ X. K' ~/ C8 I total += t# w r2 @- J: `3 n+ u) ~8 d* I/ H
update(state,t) # q d+ r6 C) E& N; r if Type[nextP]==0: # 如果下一个位置是第一道工作点 0 e" y) X- [+ I t if rgv==1: # 然而载着半成品! K$ \1 [+ \( U" j, @ Q4 Y
seq.pop(index) # 去掉这个元素并中止当次循环进入下一个循环* C0 D! G7 q! y |
continue $ J. q) a7 y) \9 h* D4 C5 m# ~0 _ if isEmpty[nextP]: # 如果下一个位置是空的+ G3 ~* z% |3 P. y# A- }
t = cncT[nextP] 6 G' E7 H' `2 r* E( | total += t / v; u3 t2 U0 y* J% R update(state,t) ! v% x1 ]& h+ m3 N6 i9 j, B state[nextP] = T1 # 更新当前的CNC状态3 u, j t0 M$ q( H* Y: M4 b
isEmpty[nextP] = 0 # 就不空闲了6 z7 B2 E) V" o' V# z$ [+ ?' M7 C
else: # 如果没有空闲5 f6 L3 n n! ], a% S1 j* J1 D- R
if state[nextP] > 0: # 如果还在工作就等待结束 % f( I% P0 v$ y g, O5 w t = state[nextP]' u. M6 [1 e I+ E" Z" g
total += t0 _5 c1 J& I& i! i' d# m
update(state,t) 3 B: z4 h% M* n7 [ t = cncT[nextP] # 完成一次上下料 8 C3 X# W2 o& L total += t1 z" Q9 V/ e4 J1 B. b6 s
update(state,t)7 X+ y: ^( N, D$ f% o* u/ c- w4 }+ w
state[nextP] = T16 F7 V: P/ _4 {9 q6 z
rgv = 1 + u% ~8 B% k9 ^: \. q9 L else: # 如果下一个位置是第二道工作点 8 K7 v5 }, K- P; Y$ h if rgv==0: # 如果是个空车0 ?( m u: q0 l! R( {+ @; B
seq.pop(index) # 删除当前节点& G4 B- V5 P" r2 ?0 _7 f6 O! P6 }0 ]
continue 1 |' j) t7 c. a/ K if isEmpty[nextP]: # 如果下一个位置是空的) R& o# O1 w% `2 Y9 T8 m' {# l
t = cncT[nextP]4 U0 E1 i* L2 I& r1 J" U0 `. h
total += t ' n8 ?9 k# D) q# _: a, ~ update(state,t)! w! Z4 F1 i! O
state[nextP] = T2 8 Z1 P# Z; p7 Y; k! R, i isEmpty[nextP] = 0 8 ?" J. W0 u! y! m& D# I# D
else: # 如果没有空闲% `/ X' c( |0 K/ }2 e! q4 `
if state[nextP] > 0: # 如果还在工作就等待结束& l z& C' L+ e1 f8 Y" U( V
t = state[nextP] ! f) R1 `4 f4 j" u total += t # C; [" L/ J2 Q8 S1 w update(state,t) & e+ ]' R- E$ j& z' Y2 y3 J t = cncT[nextP]+Tc % n9 c0 j0 W- h: h2 a total += t 0 A3 Q7 l4 D4 T9 o% U2 b% x Q update(state,t)5 X. ^3 s) K) f3 X" ^ s8 I$ d
state[nextP] = T2 6 w/ c. }# m; ]" E7 o; }' S rgv = 0 5 ?- W3 D k# \6 l+ W! l currP = nextP5 e; K U) U$ @& O8 E. D( {
temp = total 6 O# R6 L7 b& s% ]+ d3 O index += 1 * K% [( f |. _
total += tm[currP][Type.index(0)] # 最后归零 ; u/ w* B3 D7 x return rgv,currP,total: j' X6 ?# O& f( f5 B- i
9 N5 b8 D' _( U2 Y# ?3 ~: M
def init_prob(sample,state,isEmpty,rgv,currP,total): # 计算所有sample的/ s% V3 ^# c% w+ o0 C; _4 P: H
prob = []- |4 P* ^3 q+ B/ } R- M; A
for seq in sample:/ Z; W! Y; ~1 k0 @2 |
t = time_calc(seq,state[:],isEmpty[:],rgv,currP,total)[-1]# I& X& X; P( o% t* J
prob.append(t) / g9 X5 Y U4 y" k# k+ C maxi = max(prob) : O" R2 x+ q, q4 S4 V Q prob = [maxi-prob+1 for i in range(N)]0 g" P, }) P0 z0 q+ e" L
temp = 0' D6 E) i2 u1 ]& U' g+ i
for p in prob: ; c) Z2 r, a0 m) o' v( U9 D9 x temp += p5 S( C0 d) `/ V }: L2 ~, y
prob = [prob/temp for i in range(N)] 0 m" W+ X% G1 o8 S6 J! A for i in range(1,len(prob)): , H0 b$ h% n7 |0 M' J prob += prob[i-1]1 s4 B& ? \# E+ m
prob[-1] = 1 # 精度有时候很出问题( g$ X! m# h' y. U( L& [
return prob7 f& j$ b. G3 \, p5 z c# H, ]/ T
5 R, K* n5 |8 n) t' Ydef minT_calc(sample,state,isEmpty,rgv,currP,total):. K! L; }+ r1 ^, u% H
minT = time_calc(sample[0],state[:],isEmpty[:],rgv,currP,total)[-1] : H6 Q% I9 J$ c' J2 P+ {( s index = 0 + H' M, q: K9 M3 j for i in range(1,len(sample)): ; S2 r; @% N" j M7 Z6 V9 F t = time_calc(sample,state[:],isEmpty[:],rgv,currP,total)[-1]! n5 H+ q/ v" H7 ^' U% Z; ]2 X
if t < minT:* S1 ^; z1 U G$ n8 Q. p
index = i 3 O- p k6 A: r: J" a/ P7 z% o8 ~ minT = t , x C2 m" Z7 t! `& |) z return minT,index: } l: H$ `: A& |
" s4 h, _8 Q9 h; W7 I2 |def init(): # 初始化种群(按照第二道工序,第一道工序,第二道工序,第一道工序顺序排列即可)% ] P" q2 D2 v- b1 w$ }0 N
sample = []7 X: \1 D# }9 E! P+ W- d1 i5 j
refer0 = []$ `+ J+ x- G+ A
refer1 = [] 1 H5 Y% t* G8 M7 v) y; D' j for i in range(8): 4 `! K$ a, }% s9 j! Q if Type==0: ) S, @7 H+ f$ v2 z% [! ^ refer0.append(i) $ w. A: @% U1 A" c/ P. v, P else: 3 e" m" d2 w6 I- l+ V, L& N1 W refer1.append(i) % h4 P, l9 s6 [3 g' `1 M for i in range(N):( U; b+ ]: {; v! T
sample.append([])4 a' X* s7 v8 q8 S. W
for j in range(L): 4 l% ~5 S2 F" S- u/ A if j%2==0: 5 Y+ x( V o7 a8 a3 J9 V) l sample[-1].append(refer1[random.randint(0,len(refer1)-1)])# N* s, u+ R1 j# y* ~
else: / _7 y1 A7 r( A2 w sample[-1].append(refer0[random.randint(0,len(refer0)-1)])# p) O8 ? o0 u; [6 W" P
return sample ; A* `! L) T% g9 J' C/ `, D3 l( a; O5 Y. v
def select(sample,prob): # 选择算子 ; r. h: Y4 g5 V: o% k sampleEX = [] % A$ s4 f+ K; d! O0 w. d for i in range(N): # 取出N个样本; @( C0 y6 w8 x9 v
rand = random.random(); w% t2 P( @" ?
for j in range(len(prob)): x# p# n, I; S if rand<=prob[j]:, H5 D7 @: L1 W+ v6 s9 k1 Z
sampleEX.append(sample[j]) * F# T. T) B0 S) v7 b- z3 E" e break 6 Y( L" Y$ M% I! p( z return sampleEX . J% d- k/ e ^$ B+ i2 l& v8 @" o# o3 _
def cross(sample,i): # 交叉算子4 [0 l: J5 q9 _$ T5 P; f! a
for i in range(len(sample)-1): ' n6 [0 D" W u0 N, P" _ for j in range(i,len(sample)):6 X7 t; ^3 _% U( R% O
rand = random.random() 1 f ^6 G) n1 A; B+ L if rand<=croP*(e**i): # 执行交叉 % K( F; D: b( C4 {2 p' s1 g loc = random.randint(0,L-croL-1) 2 p8 u" M! S* p2 b" a6 k temp1 = sample[loc:loc+croL]6 Z- S+ v2 F/ p
temp2 = sample[j][loc:loc+croL] ; v: a+ |" S; K9 K1 A5 u: l for k in range(loc,loc+croL): 0 x- Q: X; e1 o sample[k] = temp2[k-loc] . s( F9 ~+ R1 u6 o4 ^ sample[j][k] = temp1[k-loc] ( x M/ N7 ?5 s1 m8 A" n return sample1 D7 Z3 o: b' T8 W% \* d" `( @" l
6 B- i: T5 x1 S9 Z4 Fdef variance(sample,i): # 变异算子 ) `. |2 k; D! u+ o: f" k
for i in range(len(sample)):: v1 h' O2 ?0 K6 f
rand = random.random()) g+ e3 `/ S( u0 M( F
if rand<varP*(e**i):. b0 @8 E% B$ G9 i, `
rand1 = random.randint(0,L-1)2 n) ]9 B2 @& A
randTemp = random.randint(0,int(L/2)-1) |2 I j4 R. C7 F" J4 N
rand2 = 2*randTemp if rand1%2==0 else 2*randTemp+1 ( v$ c6 a5 u4 d1 M- a. |% ~ temp = sample[rand1] ; q) N; ^: H! Z6 L* \4 X1 g sample[rand1] = sample[rand2] 5 h G: V4 T% W5 X( `5 b sample[rand2] = temp% a( V. o0 ?* A* ]8 ~
return sample) C8 d* Z7 B0 o( ~3 i1 M3 J
% Q t( s. k7 G: Bif __name__ == "__main__": l) a+ Q, {/ ?; e/ P, a' y0 d! F state,isEmpty,rgv,currP,total,seq = init_first_round()7 U8 P G7 l# W3 d$ g
print(state,isEmpty,rgv,currP,total)" r6 \2 |2 I+ U6 j/ }4 U( V
sample = init() $ Q- o/ r- i- Z" J* c+ ^8 [ mini,index = minT_calc(sample,state[:],isEmpty[:],rgv,currP,total) ' k. Q2 {' \ Z( c best = sample[index][:] ' ^1 B- M9 i+ E* W0 r7 p& S for i in range(100000): 9 F% L0 _( z, l( ~* a$ ~ f = open("GA.txt","a") ( c9 z: @" R4 k8 w. l* P5 z tmin = minT_calc(sample,state[:],isEmpty[:],rgv,currP,total)[0]9 j% H3 I, S# d; |/ J |
f.write("{}\t{}\n".format(i,tmin))" y" _8 |" v& d. _, i8 C
print(i,"\t",tmin,end="\t") ' h: \2 }' Z4 R# f! n prob = init_prob(sample,state[:],isEmpty[:],rgv,currP,total) ( q0 i+ I- Y9 P5 S, t8 } sample = select(sample,prob) ( q" x9 u9 U Q, v- s sample = cross(sample,i)) m$ t: ?* w- M( J7 r
sample = variance(sample,i) 2 Z9 S; }, R. Y8 P mi,index = minT_calc(sample,state[:],isEmpty[:],rgv,currP,total)) v( X. K3 O3 @
if mi>mini and random.random()<e**i: # 精英保留策略7 S# n+ U3 y) Z$ p! g: Q2 I$ P- v& h
rand = random.randint(0,N-1)# R3 Z/ l# n+ H/ `, Y* S! \& Y$ U
sample[rand] = best[:] 7 ^( k3 \6 K" N! P1 L mini,index = minT_calc(sample,state[:],isEmpty[:],rgv,currP,total) " k& w e6 R0 f. V* ^* D best = sample[index][:]/ ]; K/ P3 x9 w0 }' C4 T
print(best) + A/ b' u! M2 e# |% }9 u f.close() 5 P; w6 D5 C8 G% G; o print(sample)' ~( }4 Y2 H3 L! B* G9 Y, A
遗传算法这条路被堵死后我一度陷入俗套,用最直接的贪心搞了一阵子,觉得用贪心算法(即考虑下一步的最优策略)实在是对不起这种比赛。然后我就变得——更贪心一点了。+ H3 q2 Q: q: G
0 N, t7 D( V' [" W) Q! ~""" 6 i, H- P, P1 H+ h( R; j9 `d1 = 189 \: J$ y& e. I5 E* N: `2 O8 F
d2 = 32 " H d8 l! W9 ~" }/ A+ x* Dd3 = 46- ^* Y8 H" a: y5 G" \# A
T1 = 455 4 ^$ b Q. _1 S0 OT2 = 182- O9 R4 W8 z! r* x& D; N
To = 27 . R" S7 c/ j, M! y3 e8 _Te = 32 K: E( N$ c5 y, n' {) Y# G" KTc = 251 Y- T* P' d- ]$ Q
"""6 x$ U; u& B& g1 y. a" K1 H/ `, y
# B. J0 r. X4 L) KcncT = [To,Te,To,Te,To,Te,To,Te] 6 z! w9 Q4 H+ y2 E* qtm = [. ~( A) q' P" G U% l, Z1 y
[0,0,d1,d1,d2,d2,d3,d3],& a" `- d1 J: }7 d% R2 r" q/ {6 ^6 [
[0,0,d1,d1,d2,d2,d3,d3],; b; ?+ w* c1 K( L
[d1,d1,0,0,d1,d1,d2,d2], 3 o! {5 t5 }" S) W [d1,d1,0,0,d1,d1,d2,d2],; u$ i$ z i) O. X
[d2,d2,d1,d1,0,0,d1,d1], 0 E) F% q' W K: Z( ?6 b [d2,d2,d1,d1,0,0,d1,d1],5 W' {$ w* y# P. y. {
[d3,d3,d2,d2,d1,d1,0,0], - E$ {1 z3 |6 u0 ?8 z/ A& V [d3,d3,d2,d2,d1,d1,0,0], # Y2 E: S9 f' z- A] ) c: n& }' \( g. CType = [0,1,0,1,1,1,0,1] # CNC刀具分类. I( k# L3 u5 y3 ?: H% P
% ?" l- e( j; J, aA = [] # 储存第一道工序的CNC编号 8 O, p7 _6 F$ A$ vB = [] # 储存第二道工序的CNC编号 # x% y( l8 E; C' Q) G- Gfor i in range(len(Type)): 3 a' s# ^" \; M if Type:& e( p- K2 N2 \; k. w
B.append(i)4 x S2 F) E6 ?7 R% Z5 e$ x0 q
else:1 y" ^" X X; C0 G, X
A.append(i)' V; [; l1 l7 ]& k. W
/ Z; }' E5 Y' F7 r- [8 ndef 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) $ Q8 ~4 V* `3 b# ` [; E state = [0 for i in range(8)] # 记录CNC状态(还剩多少秒结束,0表示空闲) % w, }' R( U" T, g6 K) Y& J9 m isEmpty = [1 for i in range(8)] # CNC是否为空 * B0 H! `' o1 z. N log = [0 for i in range(8)] # 记录每台CNC正在加工第几件物料 4 b: P' a1 u( O' [ count1 = 0 * u1 f9 N" t. }4 A2 w+ r4 B0 A& y rgv = 0 # rgv状态(0表示空车,1表示载着半成品): S4 t4 e2 x" D9 i! w
currP = 08 t# u% r7 E% |% C3 N9 s
total = 0# c( e. J. u0 N0 @
seq = [] 0 Q* q6 D" H5 G0 [& P, X; D( m/ f flag = False* w% `; K; B- A9 b
for i in range(len(Type)):2 a3 w" Y5 I L! b3 {) x
if Type==0:' b1 u) x) Z0 G# p" g
seq.append(i)" L; N( m$ ^' J; T0 l9 f( `5 i" y
flag = True* D5 g+ E* R- ^4 a4 e5 Y1 `- x. c
currP = seq[0] 8 P( |; }6 S" B0 j seq.append(currP) 4 B0 y. @+ V3 r6 z count1,rgv,currP,total = simulate(seq,state,isEmpty,log,count1,rgv,currP,total)- c: c1 i! S( F! M) X) [
return state,isEmpty,log,count1,rgv,currP,total,seq/ W2 h- @. Y' o) Z. {
" w* q7 i' S& I* r* A J+ X
def update(state,t): + j+ \1 @5 Z' ?" c2 F3 R for i in range(len(state)):- Z' U( ?; V1 t+ q4 [; H8 H
if state < t:6 ~; p9 T. J7 m3 R5 R
state = 0 - U/ ^. p8 g) o' b5 W3 e" g else: 5 P' W6 x! ~( {! q ` state -= t/ A' ?: g7 y; }' R& ^
1 ]0 D; q# V0 i- [def 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)0 v# q& C8 z( j j
index = 0& m6 _: ^# C! [- q. }
temp = 0: N& n! X: e# J+ O! E. F- K* O8 @
pro1 = {} # 第一道工序的上下料开始时间 1 D% f( ^ y: T# Y( p6 l pro2 = {} # 第二道工序的上下料开始时间* D* T3 N4 c/ D) ?3 V/ |# B8 [
f = open(fpath,"a") 2 d; v. U2 E+ v7 b: u! m4 U: B while index<len(seq): ) y' v o# f4 b5 ^8 ] print(isEmpty) # N2 H( |8 b9 d# {3 l7 ~ nextP = seq[index]( k8 u: ]+ u! v8 K- ]& c. @4 k: R
t = tm[currP][nextP] $ C8 E4 ~2 E! q$ F5 H# n) I total += t4 Z; ^; R: ?/ Z B& J0 e- G
update(state,t)! i, N [. o% p4 N- h0 L( f$ V
if Type[nextP]==0: # 如果下一个位置是第一道工作点/ g& x! v% v# d1 K* M/ I
count1 += 1+ k! j/ J2 i& Q. P' M0 B; v8 U
if isEmpty[nextP]: # 如果下一个位置是空的$ x3 h' v/ d0 k3 t9 I0 Z8 T- M* F
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)) 0 F* ^7 e& J6 s4 L t = cncT[nextP]: F6 }7 L" i7 X0 H# N4 F9 y
total += t 0 j) T# o) ]3 Y, D [4 r update(state,t)2 I. l1 h b8 H
state[nextP] = T1 # 更新当前的CNC状态( p! W F% a$ b( @
isEmpty[nextP] = 0 # 就不空闲了 ) A. O& ]2 i6 {, a( A7 j$ M Y# i else: # 如果没有空闲7 q ~7 @' D/ A) F
if state[nextP] > 0: # 如果还在工作就等待结束 : t, g4 x5 \$ \+ r, |" Y* s* S t = state[nextP] 1 F$ \. E% i, u9 E) v) Z total += t1 ~; x0 H; W+ F0 K6 c( R
update(state,t) * J3 o; [& g9 B: r 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))% P& `" T7 n& }5 i2 l
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))0 i' t' K7 o% c+ ]/ X# w# L% q/ U
t = cncT[nextP] # 完成一次上下料 , N8 |4 V" w; o1 l) Q8 |" J total += t3 _1 O$ Z) q4 h: {2 n1 j. j
update(state,t) 8 e2 ^! |0 v2 H) @( c" i state[nextP] = T1 , p" h q' k8 a* e4 J" g rgv = log[nextP]" ~) u) H7 _. {/ Q ]9 }; n: \! B- J& t
log[nextP] = count1/ C6 ^; s# F. N2 G
else: # 如果下一个位置是第二道工作点 ( o$ ^) ?9 M) i V# I" l if isEmpty[nextP]: # 如果下一个位置是空的8 g, i0 G6 C% h) A d6 v
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)) H; ?+ u1 x% A' p
t = cncT[nextP] $ n ]5 O4 u0 I5 b* m/ ^0 F total += t / y1 G6 U3 I" p& T w update(state,t). z0 v& W: J" m( J6 a. {% x- L
state[nextP] = T2 2 W4 H/ k% n( a1 n$ ~7 l% m3 W isEmpty[nextP] = 0 3 {. r5 l, m! V4 [1 Z* B. V+ I, ^) Q% k
else: # 如果没有空闲5 Z% T8 B+ x a" ~& W/ X
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))- A/ j( r" w% [. k" m8 a, 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))0 d! A' `$ u8 G% c( N/ A
if state[nextP] > 0: # 如果还在工作就等待结束 / I# ?9 ~- e4 r& @( G$ F4 Q t = state[nextP] $ d" x% ^5 c8 _& v4 N( s total += t 3 Z7 O. y w# {* h( S update(state,t) 8 U) r. m" g6 N3 U& N9 w4 I t = cncT[nextP]+Tc/ x0 z* x4 N; q r/ v
total += t8 j7 b6 a6 q C( O& ]4 f- O
update(state,t)8 D/ O' [" [0 K" Q( Q" S
state[nextP] = T2 7 q6 A5 W' y3 N/ K* D/ m' Z: q log[nextP] = rgv 3 h4 c/ v0 m* }' x5 C rgv = 04 u6 S6 D& a& M3 ?+ E. X2 _
currP = nextP: M( b E8 C K( s" h. f
temp = total ) G7 f* m2 y6 |/ J) T a8 u index += 1 , Q- Q+ n/ _8 ` C f.close()- @4 u0 f! |4 w9 T" B. w ?1 W# J
total += tm[currP][Type.index(0)] # 最后归到起始点 _- U5 j2 z/ q0 \ return count1,rgv,currP,total " E$ X4 t! \" d! e , x+ E4 S7 j" E& ^# sdef time_calc(seq,state,isEmpty,rgv,currP,total): # 主要用于记录时间- v, R1 S) L- P
index = 0. G/ m9 r1 x6 \
temp = 0 / j9 M4 l" S( E" X- h: i5 Q" w% d; v% k while index<len(seq):, }, n7 z e6 d+ i4 i) ~8 D
nextP = seq[index]4 Z, [, q( F! m. d
t = tm[currP][nextP]. Q3 @: k! y6 |$ r% c
total += t* G/ l. s6 q8 y3 j
update(state,t)" B- d7 d' P8 D
if Type[nextP]==0: # 如果下一个位置是第一道工作点; m/ j- i& y3 j6 Z5 ^& b
if rgv==1: # 然而载着半成品 4 p( w1 }3 f- A) f9 B0 V! Q* n seq.pop(index) # 去掉这个元素并中止当次循环进入下一个循环- ~* U7 f R6 m& A& f
continue ! U( X- F% U7 I k2 L0 e) z if isEmpty[nextP]: # 如果下一个位置是空的 4 ]! E, \3 R/ j4 h9 \! n' o t = cncT[nextP] 1 g0 g* Y- X! A/ P) t& l+ R/ P4 ]+ u total += t 2 _ O* g+ k! m! V3 P- B3 s3 g, u update(state,t) ) r/ r+ _3 T2 B! [8 y; f state[nextP] = T1 # 更新当前的CNC状态 ( L( R, E! k: J6 b3 j isEmpty[nextP] = 0 # 就不空闲了" `+ v- P& q3 A. A- L
else: # 如果没有空闲3 W% G( a* ]0 r1 f; t
if state[nextP] > 0: # 如果还在工作就等待结束 2 `5 n# n- _5 d0 e* w: O t = state[nextP]' G# C7 X: n/ O5 l M8 A% F2 @( q' |
total += t6 _' t8 x0 A$ g3 J; C( X
update(state,t)$ f( g0 C5 f D9 |5 L
t = cncT[nextP] # 完成一次上下料 8 N9 L' {; b2 [. ~5 p: n1 y+ v total += t+ M6 q1 A: r; n! v6 I9 s7 _/ R, O
update(state,t)6 g H! a# [7 R! q/ G
state[nextP] = T18 _* \3 e4 _3 c. \' i. S! q
rgv = 1' U& g7 f% w! V
else: # 如果下一个位置是第二道工作点 . s4 m& d, t- U if rgv==0: # 如果是个空车 + e$ x0 n! {3 O: x7 P seq.pop(index) # 删除当前节点 ( I7 q# I$ H {) _; X# s" H k" b continue- R6 w9 `; Q. K; g, k d3 s
if isEmpty[nextP]: # 如果下一个位置是空的, P8 }* |. Q; J7 p
t = cncT[nextP] " z1 I6 e/ C$ X total += t( O: N* c3 e1 S% N% F
update(state,t) - ]. [6 \9 A8 e6 z4 {0 |( j state[nextP] = T2; W! `. \1 Q# l/ K( D1 e
isEmpty[nextP] = 0 ( N4 w2 _. o: `: y' Q1 x
else: # 如果没有空闲9 u% `# e& _$ [/ `
if state[nextP] > 0: # 如果还在工作就等待结束. b% ]) ^- \8 Y' N
t = state[nextP]( e8 `9 j" c: B, R( i3 U5 {
total += t* S; m3 { k8 R1 m
update(state,t) 4 g5 y+ w j+ x S; z/ k0 ~' @5 i t = cncT[nextP]+Tc. V7 g1 R; G& [0 n( [5 P- S
total += t* y; T! d: Z1 s+ ~5 z* Y
update(state,t)3 ?9 d! c3 B" a* E K5 a
state[nextP] = T2 2 G, b# F9 B2 u+ @! E0 O2 w rgv = 0 8 E: v& F7 d, C6 C6 m% o _8 i5 X currP = nextP, X5 j7 ^' o& J1 R" J' `# _0 M
temp = total ! l' H E! a3 B0 @% Q index += 1 0 z+ e! _6 S% t6 r/ U return rgv,currP,total " X: @3 Q2 r' K4 {7 i q& A $ [- k' i1 h7 o6 |; qdef forward1(state,isEmpty,currP): # 一步最优 - i% ~! p4 m4 O) z. e lists = [] 5 p* ~4 B! Q. _$ Q( f1 e+ F if currP in A:& i3 n2 |& Y8 ^* f
rgv = 1; B9 I4 Z: q6 y
for e1 in B:9 K8 W% U7 a2 k3 `; m" a/ _ R2 Q
lists.append([e1])8 U+ i5 }' @0 j( H4 }
e7 P. {0 p7 _) g" q4 @1 V else: ! v1 h; m/ ?) E* ?) y m1 s rgv = 0 & D/ a, R0 o) ^8 l# ?. Z) h for e1 in A: ?5 b+ @4 x% p W! _ lists.append([e1])1 m! i! \8 g ~0 L9 I2 a
1 g: ?) S, o9 k" T7 H minV = 288001 U9 F6 V) S, t3 k
for i in range(len(lists)): 8 T; X7 C1 V- L t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 4 t& _% v" h, g1 P+ w7 @ if t<minV:2 L0 U) s8 C7 E* R
minV = t # f I. U$ I) [& E- w index = i ; k1 b3 r" N4 ~! z/ R' s8 ` return lists[index][0]- j7 J( R8 R7 ^
s/ h$ \9 l3 W: `: M' S$ Tdef forward4(state,isEmpty,currP): # 四步最优! D' @, a( S9 T/ ?4 t
lists = [] ' j) z- e: S7 N8 m/ Z( D2 `3 d """ 遍历所有的可能性 """ - g" r, m1 l3 X6 h3 w( S$ v, [7 Y$ J8 Q r if currP in A: # 如果当前在第二道工序CNC的位置) Z' l6 S" q+ N a" L
rgv = 19 w+ R. {+ U% D# @
for e1 in B: 2 s+ E& c0 w! Q O* x: q/ `# {" k. P0 g for e2 in A: " q" L, u" \+ f( R2 M# R for e3 in B: p: Z8 Z# d: H/ m" a# ^ for e4 in A: * D" g i) E8 u* q: s5 @6 t lists.append([e1,e2,e3,e4])4 S: u4 ?& C4 t+ [/ P) m4 w
else: - W( P: q" X, K% O; n% z rgv = 0# \ i9 U+ k/ K9 l% O
for e1 in A: 1 A0 ]# b$ E3 X2 f" x- h for e2 in B: ) d! x" q- G( A for e3 in A: 3 p' p% S. C6 L% Q& E for e4 in B: @# O9 ^; b5 Q: M
lists.append([e1,e2,e3,e4]) . r$ A4 M3 H7 O: W# v* h" A+ V minV = 28800 y& `8 \! P# V8 Y' J7 _/ q. E
for i in range(len(lists)):; Q6 X. j' t3 P; j' j* O! _+ \
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]' F$ O, c0 b* b8 ?+ I+ T
if t<minV:! o6 h5 l3 q$ @9 x) I
minV = t ! ?* x3 h/ V9 z1 y- i index = i * X# A. g6 O4 F" d0 P return lists[index][0] # 给定下一步的4步计算最优 8 h. u& R7 l/ l1 E+ M# ~; Z' {3 H& S1 z' H2 ]
def forward5(state,isEmpty,currP): # 五步最优. `" n" _: e/ u( R$ j% D
lists = [] + h8 y" Q$ N: w3 N """ 遍历所有的可能性 """5 c7 F7 q! k& i G: f. l# F X
if currP in A: # 如果当前在第二道工序CNC的位置$ V( U8 F. _. T# `. V
rgv = 1 5 M1 w& W( W$ ~4 H for e1 in B:& E" z$ R0 E/ z6 \8 \
for e2 in A: 3 K0 z3 z. O% y* @ for e3 in B:; K# ]) p% \5 i; [" f" [) H
for e4 in A: ) N7 [" j4 c2 y" \, V8 n for e5 in B: 6 \/ R& {( R6 o; }- L lists.append([e1,e2,e3,e4,e5]) ' S8 v" a$ d8 t+ O2 d$ R else:# F O8 E0 n( T4 t) ?8 [. E
rgv = 0 % B' O/ S6 R1 k) c' R for e1 in A: * q! k' T; v3 Y4 l f for e2 in B:) a2 C. |7 ~7 ~0 j; Y; _
for e3 in A:+ h/ }4 {- a3 e9 r5 h! u
for e4 in B:# h& x! c. U, a" H
for e5 in A: ( a2 Y5 E6 H" U7 J2 { lists.append([e1,e2,e3,e4,e5]) & c& q" P+ \ r i minV = 28800 * D5 o8 k0 o, [" j y for i in range(len(lists)): 1 N$ x! _, Q9 S: } t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] ; o; ~$ P; {8 N) o$ R0 [: e if t<minV:% `7 d# E! @) {
minV = t 7 ^: q2 S) e+ x2 r; F index = i( V. t) H, ^) e
return lists[index][0] # 给定下一步的5步计算最优5 ]. o, H' M& I
# C( H7 |& S$ _% }) F) e8 Ddef forward6(state,isEmpty,currP): # 六步最优 " I+ s/ l2 ]( A9 w- ]/ e+ V! a4 G lists = []( \! J' |" P" o6 H4 W
""" 遍历所有的可能性 """ $ y- U% y8 C& E, W, r! @ if currP in A: # 如果当前在第二道工序CNC的位置 4 T. T" V; V( J2 l% z rgv = 1 1 S0 ~9 B) F6 T8 D+ s; H4 O for e1 in B:* L% v+ R; }( w& \5 G
for e2 in A:- _; M3 _+ O9 Z3 z6 ^2 C
for e3 in B: 4 R0 t1 N; v0 H, @7 h' V3 c for e4 in A: 0 M2 S/ I7 \- B ? for e5 in B: ' I9 Y5 Q6 g! r4 k: A7 _ for e6 in A:- E' `6 s1 t: ^4 c5 {4 T v
lists.append([e1,e2,e3,e4,e5,e6]) 0 Z k/ b7 I) O4 f: k4 z4 n else: 5 L- ]8 K. K" d3 t2 ` rgv = 0 ; t) E3 F I* A( ~" @# f) c for e1 in A:0 y, P6 }; A8 R
for e2 in B: ^7 A8 a5 O' h( l' W) j
for e3 in A:4 G+ w& o- F+ o
for e4 in B:5 Y) i9 Y; O3 h! ^9 o6 l7 v
for e5 in A: / ?- e: m5 v$ c$ G! B for e6 in B:/ u# ?% w& p5 Q( g
lists.append([e1,e2,e3,e4,e5,e6])0 v o, y+ i" o( R3 ?
minV = 28800% N7 ]/ {6 Z; Y
for i in range(len(lists)): 0 \, |/ L1 m: {3 H7 v( Q t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 7 ~/ Z- |+ s; n8 E) R* G3 b0 e" p% s if t<minV:+ f' @/ P2 p9 L3 H& P$ F; h9 {
minV = t ( S, i' F! ]+ j; V index = i + G- y: c" I4 ?0 f return lists[index][0] # 给定下一步的6步计算最优" q3 z5 ~4 O1 V; l/ {9 I
2 R' r2 [) K0 V9 X, B
def forward7(state,isEmpty,currP): # 七步最优" \6 ~& [* y: k% `; G0 h, {
lists = [] 9 u4 I0 B3 l. Z% C* B """ 遍历所有的可能性 """ / V( `" Y$ s p if currP in A: # 如果当前在第二道工序CNC的位置3 I# t! g& l: b
rgv = 1 ' L/ z* h: D0 @9 t, x# v3 T for e1 in B:/ y3 J3 N/ j7 X
for e2 in A:( V9 d7 k1 {( f8 H, A
for e3 in B:6 Z+ D6 `4 y+ X4 P# o
for e4 in A:& M. g2 l" C4 {$ n* T4 d
for e5 in B:0 @( a! q8 O7 a$ C* G2 x
for e6 in A:6 I/ r% U( W. ]7 H6 G5 q/ C+ j8 g+ t
for e7 in B:4 H- ~$ ~* Y# k: M6 N3 x
lists.append([e1,e2,e3,e4,e5,e6,e7]) $ u: r; Z7 W" X4 ^ K else:/ w* X" Y& o0 l1 z
rgv = 0 7 K- l& R; ~, f0 a for e1 in A: , _! e) k# h, G/ x1 c; Q4 p for e2 in B:- u" X. W% g# a# k9 X# J4 Z
for e3 in A:3 L @6 d+ k4 R/ ?6 K
for e4 in B:! G3 e" e) T _4 J* A1 [" ^, u0 R% K
for e5 in A: 2 Q5 ?2 H) J& D+ C' i" p/ w' W for e6 in B:8 p `) A, i; w. I( O' }) s
for e7 in A:% T$ j% c# j& x' i' H3 C1 }, N
lists.append([e1,e2,e3,e4,e5,e6,e7]) ( n' y8 m! y( j( m) }0 C minV = 28800/ k' W4 ?3 p9 d0 w8 n& T% d h o
for i in range(len(lists)): 5 F1 ^5 w P8 Y$ x t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 8 o/ B9 Z% z V: G if t<minV: 9 E( w* [. o) M; `2 N( W minV = t' E+ P$ ^. |* {9 `
index = i ) \8 g z' u, g- d, s: d return lists[index][0] # 给定下一步的7步计算最优6 D3 r% M/ v) R! H0 y/ a
4 B9 s* c# }! x y' P8 C5 M
def forward8(state,isEmpty,currP): # 八步最优 0 H0 }) U2 C! z0 {1 I lists = [] , W# |8 x. L( j! D+ M """ 遍历所有的可能性 """( H ~0 n5 s9 `" z3 z
if currP in A: # 如果当前在第二道工序CNC的位置& K4 E2 u# B) n! q+ `
rgv = 16 p) e5 [) B- ?
for e1 in B: ) n! p' H) d; m for e2 in A:! S! B" o! z! N+ E$ D& o
for e3 in B:/ Z+ k1 k( o H
for e4 in A: S2 P" C h% K6 N
for e5 in B: & U' u. V D5 i$ | for e6 in A: - j2 T# |9 o4 ?2 Q* @+ J% x for e7 in B: 4 j0 }* f: j; ^0 b for e8 in A: 8 s, h7 D/ O) D1 U+ O3 n lists.append([e1,e2,e3,e4,e5,e6,e7,e8])& g9 G7 m& P) b% a
else: - g3 ~; i6 ^/ N5 W' R' \ ?6 w rgv = 0& [1 N6 Q9 T( s# ?) W& |% b9 g* K
for e1 in A: 7 e+ R7 S5 `0 E: Y% A for e2 in B:. _# [1 v! f+ v( s6 ] E) j$ v
for e3 in A: 3 B9 Y8 H( t8 R4 U: ^) O. \1 F for e4 in B: ; ?6 Y# U- ]- Z2 P for e5 in A:# y; R# F+ v |, ? c& E
for e6 in B:8 b E8 t" E* A2 t5 a! z- {
for e7 in A: 5 G. g& w& ^: I, O2 l for e8 in B: # H5 o) I7 x- }, I; r; i lists.append([e1,e2,e3,e4,e5,e6,e7,e8])4 z7 B. q4 O3 n& f+ R
minV = 288001 G% L' s0 \- g
for i in range(len(lists)): ; o8 K7 w0 g2 X) i% u t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 1 i" P, W! X; a if t<minV: . n7 x) G3 n" J/ R5 a4 O minV = t # L( k, h, I( g- m* i" t; ~2 c j index = i7 l: L" X7 x0 [; r7 n, n3 C
return lists[index][0] # 给定下一步的8步计算最优8 T$ H/ K J9 z& v6 n
. `2 O1 v. z4 c/ k1 zdef greedy(state,isEmpty,rgv,currP,total): # 贪婪算法. {! v0 h! @% q0 `
line = [] 4 r) @' D' i% H count = 0" ?5 K5 f2 J* X& D
while True: 0 C2 g" w, H" j) O0 C #nextP = forward4(state[:],isEmpty[:],currP) $ O. P& \4 J$ m4 y6 E6 D! A nextP = forward5(state[:],isEmpty[:],currP) . n: s- C& _4 g
line.append(nextP) : w6 l- S) S5 F4 T# ^ rgv,currP,t = time_calc([nextP],state,isEmpty,rgv,currP,0); d2 [1 d9 C" G
total += t! M: J& _ H4 x( u4 g7 I
count += 1 , q0 D! u$ G k# X: l& i if total>=28800: 8 |, a# l9 W8 z9 Y break! m; t: H0 R+ a+ n0 X6 }$ c( A% l
return line 6 v) k8 w7 t: C$ |7 f3 [ 3 @+ s4 H& j( n Lif __name__ == "__main__":5 ?0 y- }, o1 B$ L& d S6 L" V8 S
state,isEmpty,log,count1,rgv,currP,total,seq = init_first_round() / Y* f% E ?6 Y4 B print(state,isEmpty,log,count1,rgv,currP,total,seq)4 E r; x9 t7 }. P. Q
line = greedy(state[:],isEmpty[:],rgv,currP,total)( F1 h" d, i7 h( W' H6 E( E& w4 c
simulate(line,state,isEmpty,log,count1,rgv,currP,total): W; U: W( }) i7 y. U% [# U3 T0 v$ D
( E- ]0 X ]' R& h5 n! M$ P8 c write_xlsx()% q$ _& F: j, \2 S
后记 * q9 k) c/ @5 Z( Q. W& \( n" W6 s5 X6 j& Q4 i
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!+ m% r! c/ h; q$ m
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