; P0 k3 \$ p$ i4 |3 Bd1 = 23 # O3 z) j' O$ md2 = 414 t) |4 k! u+ K7 u" ]
d3 = 59 0 I, E9 s" G8 B/ i6 `T1 = 280 : U1 s; Z/ h7 x7 U5 }7 jT2 = 5009 N( x: u6 U [
To = 30# W3 r: H( Y+ C0 k/ }# D6 i" s
Te = 352 @: y! P P, `& G3 T
Tc = 30" a6 X% K: P; F; j( f' x
0 P6 ~) ^% Q t! I
+ B: B i% ~( f9 e5 B9 m6 j# 第3组6 u' _0 ^, B$ B5 \ H* c
7 l7 J! ^% F! z- A""" * V- S5 n' H* B6 g3 \6 xd1 = 18 / G2 E9 X: U& y2 Vd2 = 32 , R3 E5 V" V+ M( R. E' cd3 = 46 8 }/ l' ]* m5 H, N( [4 jT1 = 455 1 e! i( Y5 }: qT2 = 1822 w5 @! q7 A1 Z0 j. |
To = 27 : e5 ]. J% C) H5 E6 e; F) ETe = 32' I6 u2 P4 h+ y- H- X
Tc = 25$ N/ X8 n# |; C% w! g6 o
"""# q: v# a0 g+ w) ?6 l7 h- t
4 f6 J9 H* f' j- R9 Y7 IcncT = [To,Te,To,Te,To,Te,To,Te] / D, L: j( X7 P' otm = [# G# H4 H1 W, c8 g2 H$ \) m
[0,0,d1,d1,d2,d2,d3,d3], # `' }" l' x& z" u8 l [0,0,d1,d1,d2,d2,d3,d3], : _3 u7 K: a1 p) F [d1,d1,0,0,d1,d1,d2,d2],+ Z1 x) h7 X" f2 `( \% C; A1 V
[d1,d1,0,0,d1,d1,d2,d2], 1 j$ c) ]5 q/ c0 H; v8 i! S [d2,d2,d1,d1,0,0,d1,d1],9 L6 W. M$ a$ N) F" P" z6 |. }. j
[d2,d2,d1,d1,0,0,d1,d1],: e0 I- Q3 }" c3 y K
[d3,d3,d2,d2,d1,d1,0,0],, t- r* Z! a/ K8 n# }" [( _
[d3,d3,d2,d2,d1,d1,0,0], % j2 p" M" E a6 a0 K+ z V]7 T& F: C3 u" m7 o/ ?. `9 K# s
Type = [0,1,0,1,1,1,0,1] # CNC刀具分类4 h0 u" B. J" O5 i7 ]
# S, o2 F+ _2 BA = [] # 储存第一道工序的CNC编号9 S5 i9 s' w# P9 k* S1 P
B = [] # 储存第二道工序的CNC编号5 g% ~7 K1 U( R4 o3 w) B
for i in range(len(Type)):9 c- d! U! P0 r! b: k
if Type:3 [6 \/ Z) J! O. a$ ?3 Q( Q
B.append(i)0 c8 P7 Q2 V: i/ t
else:5 o7 |+ H% C: N- c
A.append(i) 5 [" t% F1 [, K1 T. m: \0 S1 D8 M' W( @2 w
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）. k4 W6 f: A' D+ b) o1 t& {- {
state = [0 for i in range(8)] # 记录CNC状态（还剩多少秒结束，0表示空闲） 7 Y/ M. b: ^/ P isEmpty = [1 for i in range(8)] # CNC是否为空 ! F. c- E! j: E7 K% f: [& u. L8 q9 y* x. N log = [0 for i in range(8)] # 记录每台CNC正在加工第几件物料 ' ^( [) O9 a8 d" Z7 U count1 = 0# x1 D! Y) w% S: ^8 I6 N
rgv = 0 # rgv状态（0表示空车，1表示载着半成品） % Y8 U6 X4 k% J currP = 02 a. s4 W5 |# I% Z5 g6 U3 e
total = 01 h0 ^! ~* M& I6 K7 k
seq = [] ( J0 D8 H/ |& d* R+ V flag = False : s- }7 u' i2 K6 K$ u K; { for i in range(len(Type)): 3 \( S% |2 ?- q if Type==0:7 t- d ?% F' k( a) J% U1 b
seq.append(i)% K% t3 k5 ?! b
flag = True1 e- F7 V/ t- [) L. G9 a1 z! o0 I
currP = seq[0]+ v5 g& E e2 Y- e$ N9 u- o
seq.append(currP)4 f) z# ~, M' o' @
count1,rgv,currP,total = simulate(seq,state,isEmpty,log,count1,rgv,currP,total) 6 j9 W @9 H; G5 l return state,isEmpty,log,count1,rgv,currP,total,seq4 V! f7 @7 J" b q }
) S6 Z1 r c4 W; l" G
def update(state,t): 5 b4 U4 G# c* s* d7 t for i in range(len(state)):9 h7 k: Z/ T% T/ o) q* f
if state < t:- _* `& t1 l! J. A( p
state = 0' N& o* @# A8 H6 j' S
else:6 x% }( r8 m6 ?4 b% M( V" B: Y
state -= t6 @4 B; b8 s: o: p/ t% O
3 H: g3 z- R- i, J; Q# E% W0 {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）' f/ B" P' u" W; d
index = 0 $ I' N6 y$ ~9 F$ a C ]8 K* T temp = 0 4 N- k9 j' H$ Z pro1 = {} # 第一道工序的上下料开始时间 ) Q4 Y0 W% e( ?; J! l8 J pro2 = {} # 第二道工序的上下料开始时间/ d4 A0 ?% [9 a) s/ o
f = open(fpath,"a")' C1 ~. Y+ Q8 M5 z n. \, \2 `4 m
while index<len(seq): 1 j; H; C k& g+ j print(isEmpty) 6 C; Z) u' L5 J# d nextP = seq[index] ' f: }, L5 S9 C7 E( V7 a2 p: X t = tm[currP][nextP] . w$ r8 }6 t7 {1 ?/ I8 c" v total += t/ b, E p. i0 C9 r o# m
update(state,t)' U3 ], o* p1 R
if Type[nextP]==0: # 如果下一个位置是第一道工作点 5 E( o3 b) S3 G" K9 j$ O* i count1 += 1( Z) p) Z& z: X3 ]8 `% S1 r
if isEmpty[nextP]: # 如果下一个位置是空的! ^" J: s( o+ S
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)) / E7 U& b. G5 {$ `/ k8 f, I7 G t = cncT[nextP]4 K! M' j. G/ f' I
total += t % P* b, r% r/ a* {- y0 K6 G update(state,t) ! H$ ^% @+ Y0 N: D$ O state[nextP] = T1 # 更新当前的CNC状态" ^+ Z2 o+ y1 N/ G/ j
isEmpty[nextP] = 0 # 就不空闲了: D* N) G* M& s1 }1 O# _* u" x
else: # 如果没有空闲 " |7 W' T; W9 g g( G7 M if state[nextP] > 0: # 如果还在工作就等待结束) N1 f9 _+ _1 v2 }8 Y' f' b
t = state[nextP]! l7 Z1 w5 M0 B% ]
total += t! w1 x8 J$ g* U: H6 U; m9 h. {" `
update(state,t)$ y" X$ \; m; B
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)) 2 w4 c5 ], l) A1 p. M0 K" a3 K0 Z 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))- k: m, A6 y0 C5 d) y9 R
t = cncT[nextP] # 完成一次上下料 / \- @: G5 H0 G, S& e total += t - ]2 d# G/ g4 {+ x) [ update(state,t) % }3 @ \# E. y* K- ~% b1 M4 Z state[nextP] = T1 # Z# ]" u# |; s1 l rgv = log[nextP] 3 r) `0 @+ |+ E) s log[nextP] = count1 ' A L1 A, X& L6 E6 [, V9 e else: # 如果下一个位置是第二道工作点* F4 [& ?) s/ h) {
if isEmpty[nextP]: # 如果下一个位置是空的9 j% D3 Q' i- ]7 O8 c8 `/ {
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))# [) ]5 o; ?7 H# v- Q/ e! u# P4 F" ]
t = cncT[nextP]. _" {5 Z* d1 p* T( a
total += t# K, B$ L- e3 d9 p. Z- ?! I: Q
update(state,t)5 Z! j# R7 x$ J2 J& c3 q
state[nextP] = T2 / E; b& L* ]# q; o4 m isEmpty[nextP] = 0 5 l4 h5 V% E, j f
else: # 如果没有空闲" O( Z1 |! i0 l3 w6 n( |: l, I! G
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))# D: `" J9 d- q0 _
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)) 8 Y! `" O. C' [& ` if state[nextP] > 0: # 如果还在工作就等待结束 h6 e; U: n- N& c t = state[nextP] & J9 l, a* \& r6 N2 f total += t# O6 k+ n: R' K3 T. s* |
update(state,t)8 [! `/ W' a( I) u4 i& F! x, i
t = cncT[nextP]+Tc$ G/ e, F' P# V
total += t 3 c$ V& f6 b$ U, @+ P3 T update(state,t) # S+ q& u+ ?4 h! s- j1 K$ q state[nextP] = T2, G% ?5 s2 e( T+ m
log[nextP] = rgv! J5 U- H; v L) G5 G( o- h
rgv = 0 9 z/ [" Z6 p! Y8 S" O# y currP = nextP % E$ n: }$ {9 B9 H/ X temp = total + e) V: ]# T! h: I# V0 I2 d
index += 1 ) h ~6 V# V+ R/ l: G f.close()3 Z: B k2 \% f( _) h) S
total += tm[currP][Type.index(0)] # 最后归到起始点0 q' \) s- Q) n$ S
return count1,rgv,currP,total, p1 P0 x3 l0 L
4 P( d, o C3 @+ B% I
def time_calc(seq,state,isEmpty,rgv,currP,total): # 主要用于记录时间$ s0 T# i% O$ l3 E2 @2 B5 G
index = 06 ~ O) c$ C1 G$ Y# P# k
temp = 0 , M+ @ }9 Y& \+ v: F! L- T; I' V while index<len(seq): 6 t! ]1 q9 N4 i! m nextP = seq[index] ' C! }! J0 o0 @8 Y: U( u t = tm[currP][nextP]2 z/ b) V( O5 n
total += t. u& {! K4 [$ r4 I
update(state,t) / ?- W A& P7 ^+ t& Y5 I if Type[nextP]==0: # 如果下一个位置是第一道工作点 + ~5 _9 G: W2 Q( j1 x if rgv==1: # 然而载着半成品 " l+ y1 }. l- b" H6 w seq.pop(index) # 去掉这个元素并中止当次循环进入下一个循环 L. g) E7 r4 x \% {
continue . a( A$ E9 n3 ^5 z, ] W) h
if isEmpty[nextP]: # 如果下一个位置是空的 9 q' ]4 x. t2 J5 l0 f) v3 P t = cncT[nextP]3 u# x, @8 n; n7 V, N* Z
total += t * {4 [; X" T' P9 R% e! f update(state,t) : D" b" J5 |0 r0 K( _7 h state[nextP] = T1 # 更新当前的CNC状态 i# K2 P' r2 i/ ^ isEmpty[nextP] = 0 # 就不空闲了! _0 `$ w+ O) ^' _$ O8 U9 l* m8 P1 q% h
else: # 如果没有空闲 6 c9 H1 r9 ]5 f1 F- N( D- Y if state[nextP] > 0: # 如果还在工作就等待结束 9 ~& C% ^$ l5 ^+ q8 [4 @ t = state[nextP]0 \2 E2 ~* X! s
total += t7 g& f" W3 Y2 C8 O8 u( H1 r
update(state,t) - X' E! a* \' C$ j9 ^' g t = cncT[nextP] # 完成一次上下料+ A4 Z$ m* l9 ^: T2 p: u8 K
total += t- w: r0 t9 m5 }1 I2 A5 h; C
update(state,t)$ @8 C/ S2 n( l! }9 K
state[nextP] = T1 * j/ a) D; x+ ?/ a( [4 A rgv = 1- Z8 H1 y# Y, W1 q; f
else: # 如果下一个位置是第二道工作点+ ^7 O4 [- d' L( |- t9 [, Q- C
if rgv==0: # 如果是个空车 l2 Q" B% [9 T. N3 v+ ^
seq.pop(index) # 删除当前节点+ |7 R* p0 W) S: u. q- Q
continue - T8 q" t/ E) R3 H+ s5 h# r if isEmpty[nextP]: # 如果下一个位置是空的 ( i* q0 y/ h- r- c0 q$ h t = cncT[nextP]7 v& S$ I" ]' f& [/ n
total += t ) q) e& x& f/ E a7 z8 D3 s update(state,t)/ @$ O/ y: Z, F
state[nextP] = T22 }5 R5 w+ C$ P# @" e2 |
isEmpty[nextP] = 0 ) Q1 e& b0 \* P! h; ` else: # 如果没有空闲: w- S/ C, U9 Y0 E5 \ o5 d
if state[nextP] > 0: # 如果还在工作就等待结束7 w2 E! |* |$ `: ]& o( O
t = state[nextP]% X* o' P! A# n# X' f
total += t% R. X6 {. T! [6 B) C, _( B3 L
update(state,t) ; d& x$ c. E& ^" V6 A t = cncT[nextP]+Tc: y6 n" B9 J) V# A: T. b
total += t8 h" K `; t$ D1 V
update(state,t)9 \4 R# m+ I Z
state[nextP] = T27 Z; H% _5 n2 M' y; A. |
rgv = 0 7 z' V! q, @6 Y- T; l; t6 n6 q currP = nextP. S' ^, G6 {5 @6 G6 ~! ~& `# t- R6 p
temp = total ; j% }2 g0 E* N* D4 F: T+ ^ index += 1 O4 }4 U5 \. p6 r+ ]% h) P1 a' M
return rgv,currP,total- [/ N, C- j( F3 {0 L$ }
c" E6 R- K1 Z" f
def forward1(state,isEmpty,currP): # 一步最优; N% d0 C9 F. l2 @: k3 d
lists = [] & @( x+ B$ q- J8 n2 {9 y! m if currP in A: , u H5 ~& t: F, x rgv = 1 7 ]4 Z/ ?: b z! w# R for e1 in B: 4 M3 R" O" _) L8 M" d8 Y lists.append([e1])/ l; ~8 z6 R' ?- L
. n. y( ^7 L4 B3 j, e& N3 O5 w
else:" b; B3 f5 H/ u5 g8 e
rgv = 03 E; v0 r4 e% g6 B4 Z6 q
for e1 in A:7 u3 U L' i+ y+ A
lists.append([e1]) $ f$ W2 P- _) I7 O- T1 R2 @ 8 i8 y6 V5 t* O/ T8 @5 V3 ^8 e/ U/ t minV = 28800 1 F" S. h, s# ]; d/ m for i in range(len(lists)): 0 Q/ b- ~; Z' {0 [0 N+ w# D+ m t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] ! e3 ^+ D' |" t* M1 w* o; t% T' @8 ^ if t<minV: 4 k3 s. ]: J2 R) _: T minV = t: y/ A/ K3 i# J+ b0 z" `/ U0 {( W
index = i* c$ [# V+ K+ g5 z4 k
return lists[index][0] ' g* s: {% {+ K& g7 b S" F6 g( N: w+ l( p4 v/ G3 y- p
def forward4(state,isEmpty,currP): # 四步最优 6 {7 D- i# ^$ X lists = []) f# A7 b% F* g% y
""" 遍历所有的可能性 """: n M' e8 n- ~+ d
if currP in A: # 如果当前在第二道工序CNC的位置 . t$ C" n4 @. N* t2 m( P rgv = 1 / \- E, w$ [. G0 W& |8 v for e1 in B:) G+ Y" @% o( M: U' K
for e2 in A:- X) I; s7 c7 [6 r* W# i
for e3 in B: + {0 h' q: S3 C R0 R8 t for e4 in A:1 ]7 x4 S1 f: }9 z1 }
lists.append([e1,e2,e3,e4])9 q' @" q. y( j! |
else:' ~/ r) W) ?" \, `
rgv = 0 ' Z" s- @- n0 Q for e1 in A: , L+ Z$ M$ M5 v for e2 in B:; h- }" F1 M) A b
for e3 in A: 3 z; B2 N8 O) j5 w, t for e4 in B:5 @8 F" D; j" u8 S4 A
lists.append([e1,e2,e3,e4]) 9 J7 N( Z# N1 r minV = 28800; k& F& T6 X6 g8 @
for i in range(len(lists)): . F' X$ i- q; x% n t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 4 |! }& }3 [0 y _7 G4 m if t<minV: ! H; r9 ~' P( N) E: f, N' F minV = t ' h& s4 f- a; C0 }/ r+ @ index = i# r* Q9 }# {3 K
return lists[index][0] # 给定下一步的4步计算最优 V. K6 D h8 ^0 f0 L/ Y d! J) J& T ) J ?! B+ |) c; J9 B$ o S n( `def forward5(state,isEmpty,currP): # 五步最优+ }3 a' Y: ^3 O
lists = [] % l( U! n3 j( k) J """ 遍历所有的可能性 """8 K9 U3 [) X6 y
if currP in A: # 如果当前在第二道工序CNC的位置 4 E; d( n. ^ y( ]' @" Z rgv = 1 # W9 V! x% U6 _) @( c for e1 in B:% t' L1 f7 y3 D: g. Z
for e2 in A:& F; E" i5 ?6 i: L
for e3 in B: " L$ `" I' r- e7 e% c8 P0 A for e4 in A:& n, O: B6 g3 f) e5 I0 d
for e5 in B: + y( n! F1 m, r) b) I: ]/ X0 Z lists.append([e1,e2,e3,e4,e5]) + k7 z& U. L2 c7 [9 ] else: : G7 Y3 S e9 `( ~ k5 w rgv = 0 9 r& s% H$ N( w2 ]4 R2 G# S, k for e1 in A: F; k# h( H8 q for e2 in B:$ q' }# Z) M, w( ~) \" h2 n
for e3 in A:6 P6 N- d3 ~% _( w
for e4 in B:7 t3 G: h" n! V/ u
for e5 in A: : K' Z9 P' W% l lists.append([e1,e2,e3,e4,e5])2 i, G- ], E- G. i- Q9 f; E
minV = 28800 3 R6 x* [7 y! j9 e* }6 D for i in range(len(lists)):3 H# S1 r- T* Z( x/ T
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1]" |, ^3 E* a) l& \* I" P
if t<minV: # l& T2 O+ p: e minV = t ! N2 [) _6 |3 T3 ~- y index = i/ Y2 Y/ U' ^1 S: {6 L* j
return lists[index][0] # 给定下一步的5步计算最优" i; P5 T$ o3 Y2 }- @1 {& J8 v* Q
' l% Z& v2 c1 D: R" v4 Zdef forward6(state,isEmpty,currP): # 六步最优: _) M, m3 s. T" f7 b
lists = [] 4 F' M8 l5 L' ]& `: W """ 遍历所有的可能性 """ " ~- N0 l# H. l8 \ if currP in A: # 如果当前在第二道工序CNC的位置( d G- a2 v F- w" J
rgv = 1% R" g: v, _- l0 o, [% Y6 `2 N9 c
for e1 in B: 1 z: N) b% C8 i7 m; h: ~( K for e2 in A:% C* k( F$ ]8 Z4 Z: g" J, \
for e3 in B:# ^! P8 i" c( O e* f# J
for e4 in A:3 ~* i- A: I# o
for e5 in B: A' E( h- @+ y9 U ]$ {3 P' z for e6 in A:$ |/ `" D4 F' `$ Z/ F0 N. S
lists.append([e1,e2,e3,e4,e5,e6])" F+ _4 A/ `) T9 W: x& \3 {7 Y
else: . M8 u$ B4 r: j1 U2 W5 L rgv = 0 * r& R# I0 h- g, K% |5 ` for e1 in A:, H9 U0 H: j: k0 r+ B' a2 g/ }
for e2 in B: 6 b ~5 E; F7 h for e3 in A: , @9 x0 s7 S/ F* o for e4 in B: + w: V% o9 c" O3 g8 ? for e5 in A:; Q( M! @& a D' v5 e( x1 [4 z! c
for e6 in B:" ?" H5 L$ f: s+ E/ m
lists.append([e1,e2,e3,e4,e5,e6])5 l* m9 ]' ?& v2 p, o
minV = 28800& ?( w4 `2 j) Y$ y
for i in range(len(lists)): $ G$ X D* U9 K: q: G6 C, ^ t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 2 B; |4 }2 N2 E. u4 I+ O. b if t<minV:; m3 r! m+ v3 g% t' i
minV = t* l |1 ^* J0 }7 @# |9 C$ m3 p
index = i , X" `/ j6 P/ m+ Q6 _ return lists[index][0] # 给定下一步的6步计算最优 + g: n4 h! y. ?# U8 ] i1 @6 _. K& c! L- S) G2 K) M( R5 Z
def forward7(state,isEmpty,currP): # 七步最优 7 o5 f9 M- x- ]/ {9 c1 s N lists = []9 ?) W4 g( l$ n4 r0 }- O' ]
""" 遍历所有的可能性 """ 1 m' `4 h1 I" R- ~ if currP in A: # 如果当前在第二道工序CNC的位置4 c% R, x" }- O
rgv = 1; \) D/ ^( Q+ Y9 d+ s; {3 I
for e1 in B: p# p) x4 y5 ^% _* V
for e2 in A:- |, A3 X2 n. {1 C8 s$ @& P
for e3 in B: - U+ a$ R! g5 H: H( v4 J for e4 in A:$ z: E1 {2 |$ {
for e5 in B: 7 I( c% v5 T. V e' | for e6 in A:% |- Z: i! ^' H+ W3 Q* j% e
for e7 in B:( |" ^/ \& B$ q" q; g9 O4 `
lists.append([e1,e2,e3,e4,e5,e6,e7])3 Z2 q) U# Y% T# e, @ O
else: * n* D$ i+ V h rgv = 0 8 I) v* C9 }- x2 v for e1 in A: # a8 O+ w9 U! ~% b: a3 Z for e2 in B: 6 @; z1 @8 k: M [ for e3 in A: 0 z H3 {/ X2 x! S/ S V- m! U- ^ for e4 in B: ' L, ?3 {8 b2 ` for e5 in A:9 _ G" K! H( B& Q
for e6 in B:" J/ g) `" Y2 D- H X9 h' Z
for e7 in A:! y: p" _9 V$ {/ D
lists.append([e1,e2,e3,e4,e5,e6,e7]) % y5 |, d: o5 d$ T, `4 r1 e. C ?3 c; i! e minV = 288006 F, B. A, O/ j
for i in range(len(lists)): 8 r6 p" V4 X/ N: r0 A, t, J! [ t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] s: C5 j- B) v1 q/ ^ if t<minV:* x/ W r$ ^3 h. a; [
minV = t # x( i; b! E5 B7 U+ F: ~ index = i8 s; n1 D0 c% t! |' r
return lists[index][0] # 给定下一步的7步计算最优 5 z9 W2 B) ]0 R; w9 f" w 1 n* i! y: y1 x zdef forward8(state,isEmpty,currP): # 八步最优 " Q) u$ i7 m5 H! \9 H lists = []8 T0 |4 L, l& w% x) G
""" 遍历所有的可能性 """ / X" r8 p, F* A+ Y6 j2 Z if currP in A: # 如果当前在第二道工序CNC的位置 ) X' R7 v7 h9 A rgv = 1 & y4 }) t% Q: ]( V h: t) B for e1 in B:. n. Q3 b+ g q$ ?. U" q! K
for e2 in A: }* E- r5 \- g' N; Y" ^, ~% H for e3 in B: 4 c8 d( S% K; @4 w! k* ^% O3 } for e4 in A:( N5 O6 \7 Z E+ H) A& H. V
for e5 in B:* V/ o* ]' s9 R& e. _( w% u3 `
for e6 in A:& c6 G8 w# B5 K# E f3 Y; v" D
for e7 in B: 5 Q5 _/ b; ~0 I. ^/ [/ T for e8 in A:$ m( `) z6 d$ C {' p, G# R9 D
lists.append([e1,e2,e3,e4,e5,e6,e7,e8])3 M7 v F. v' u, v
else:+ a) f/ X& t! M1 A6 `" K
rgv = 0 / F8 o4 P+ R. F5 _ for e1 in A: 5 w3 @$ S& R6 [" \% Z$ { for e2 in B:; H3 ^1 g, J) K7 M
for e3 in A: 5 k; @/ u1 H! B, [# b- b for e4 in B: 5 q8 i& w2 ^+ D3 E# a! ? for e5 in A: ) F3 N) F: E8 }5 | for e6 in B: ) n4 l+ U$ G, [4 [8 `3 M9 [# H) N1 ? O for e7 in A: 7 ]# \7 g/ ?8 G5 t4 Y for e8 in B:5 e. `) i3 Y0 L5 A; F( j3 R
lists.append([e1,e2,e3,e4,e5,e6,e7,e8]), }# ~( G8 G6 Z
minV = 28800/ W& _0 d" h8 K. u! B2 Z1 n5 e
for i in range(len(lists)):+ { W/ H. |3 j- a. S8 k
t = time_calc(lists,state[:],isEmpty[:],rgv,currP,0)[-1] 3 y6 I- ~ w7 |- ?- @/ D3 M if t<minV: 4 I7 R9 c9 K0 |/ z0 G/ q minV = t " F2 e, a1 _8 E! g( ` index = i3 G5 \' G' K d8 a$ I
return lists[index][0] # 给定下一步的8步计算最优 7 L, |3 c+ t# D) @( y7 {) z: o- @, U, Z5 r. Y" \6 n% l. E9 ?
def greedy(state,isEmpty,rgv,currP,total): # 贪婪算法) X$ e9 \: k3 W" t
line = [] ' v0 I7 h# Y$ X/ W count = 0 ) e3 T9 t/ \* K& I5 T9 P, h# z while True:& U; _3 Y$ [. t& ^
#nextP = forward4(state[:],isEmpty[:],currP) , E& P/ c! v7 o nextP = forward5(state[:],isEmpty[:],currP) - l/ K( J. w0 j7 M
line.append(nextP) 2 V; s9 z* C" z: r- P rgv,currP,t = time_calc([nextP],state,isEmpty,rgv,currP,0)5 I' r+ F6 g! `8 J9 r5 G8 F) B6 @
total += t * z8 W' `" a- ?: @ count += 1 ; D5 b0 S5 v' R7 Z _ if total>=28800: + z& o- x# W) v, ~ break ; I3 t9 ?+ x! g1 z8 v return line 2 V/ g+ K! I% h) s % V' s$ {, g" L5 ~if __name__ == "__main__": ( v: F1 e) J$ T# {! ]- c6 C state,isEmpty,log,count1,rgv,currP,total,seq = init_first_round(); u2 B+ v2 T/ F" |
print(state,isEmpty,log,count1,rgv,currP,total,seq) 3 e/ q0 \' Y8 M9 M6 J line = greedy(state[:],isEmpty[:],rgv,currP,total). M* k! ]- I1 k: C
simulate(line,state,isEmpty,log,count1,rgv,currP,total) + G# a3 o3 _1 P. I/ ~ " O/ w! }# G: \; [. r; B
write_xlsx() ; w1 Q3 l4 z$ Z. n# R' t% F后记0 Y8 E0 q% K' V0 ~
: U! S: M% @6 c9 R% D* q" |! V
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！. X& y/ Y% F, z
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发表于 2019-4-12 16:18
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