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发表于 2009-8-16 18:59
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修改一下循环的下标可以使得程序编译通过,但是不知道是不是和你的模型相符合,你仔细分析一下吧!
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- model:! ]& E& J: ]1 L7 _% z) O, }1 }
- !n个货车8收点运输问题;1 A\" P) C7 d+ V
- sets:
) Q; w b1 g0 z* V+ D0 R1 j - place/place1..place10/:time1,time2,timexie;0 ~/ g# H, E' T, E* O- W: V4 F. p
- car/car1..car3/:time;
8 l- j& U' a0 a' D0 k - 4 H. \ A2 D8 I
- jiu/1..9/;8 @4 J! g8 U; ?0 o. g3 g/ J; g& S
- er/1..2/;
9 E# [% k7 P( V6 k( W
+ ?/ z0 U. V, B! t+ K$ z- lc(car,jiu):lucheng,ct;
; Q% D) Q4 @7 N) U* w U3 v - plan(place,place,car):a;
( ]# w* Y) u# X4 Y - distance(place,place):dd,dt,which;
0 K% K$ u. T4 {3 V6 L7 x\" [* y7 o% ? - endsets9 ?9 X! p! y\" H. Z5 S- ~
5 F: {$ u+ n2 N9 _
9 @( n* q* p3 P, b( C8 s# T- !这里是数据;& s8 p) [+ Y% p$ X8 T
- data:
3 d9 M$ O4 T$ @- q) a6 ~: T - s=1;$ m5 e) Z/ }% A6 e+ W {
- time1=-10 1 4 1 4 3.5 2 5 1.5 1; ' h! n& ^2 r: v* K
- time2=7 4 6 2 7 5 5 8 4 10 ;! o: h' {2 _0 p' w$ ?- b
- timexie=0 1 2 1 3 2 2.5 3 0.8 0;
- [4 l' t0 x# |6 ?! l
. y. g( u& T/ t( N6 R7 p8 o0 F4 l5 T- & x6 Q, M) S! K. `$ C
- dd=
) a' k: H8 H) k+ G\" Z6 @ - 0 40 60 75 90 90 100 135 80 0. m$ ?+ X' K- W; R! L3 w\" c
- 40 0 65 40 100 50 75 110 100 402 I# b9 f; W4 N* G\" e
- 60 65 0 75 100 100 75 75 75 60
4 @9 Z% {5 e0 q$ X+ R$ l3 r. c - 75 40 75 0 100 50 90 90 125 752 M6 Z, Z- J\" z9 \2 h& [; Y
- 90 100 100 100 0 100 75 75 100 90# d: B4 V& h( p, k' t+ b
- 90 50 100 50 100 0 70 90 75 90+ e6 O: a+ v# j. j; q
- 100 75 75 90 75 70 0 70 100 100
. N7 ^6 n8 G' Y' \# l7 E9 L& a T# l - 135 110 75 90 75 90 70 0 100 1359 P0 @* F! V4 _) c2 V( ^
- 80 100 75 125 100 75 100 100 0 80( m5 D* g$ j3 }; g: g: U9 `
- 0 40 60 75 90 90 100 135 80 0 ;6 K+ c+ R D# O2 `7 p' J
- dt= 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 0/ ^. r+ z# n' Q/ b
- 0.8000 0 1.3000 0.8000 2.0000 1.0000 1.5000 2.2000 2.0000 0.8000
: C* ]: c. a4 ^2 ~# o. U - 1.2000 1.3000 0 1.5000 2.0000 2.0000 1.5000 1.5000 1.5000 1.2000( `' `6 y* n' W6 q! B
- 1.5000 0.8000 1.5000 0 2.0000 1.0000 1.8000 1.8000 2.5000 1.5000
3 I0 i% W\" h( O! h8 g/ x* o - 1.8000 2.0000 2.0000 2.0000 0 2.0000 1.5000 1.5000 2.0000 1.8000
, D* L; ?* c6 A* I7 Q/ |$ X - 1.8000 1.0000 2.0000 1.0000 2.0000 0 1.4000 1.8000 1.5000 1.8000$ i; ~3 _# u1 H o( v2 X, m
- 2.0000 1.5000 1.5000 1.8000 1.5000 1.4000 0 1.4000 2.0000 2.0000
, u8 y: ~4 N! o+ p - 2.7000 2.2000 1.5000 1.8000 1.5000 1.8000 1.4000 0 2.0000 2.7000
\" p- E4 s' G; l$ j - 1.6000 2.0000 1.5000 2.5000 2.0000 1.5000 2.0000 2.0000 0 1.6000
* y& S- Q\" Q6 A - 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 0;
: v. H\" i2 Y8 E. ?' j0 |* ^4 v - enddata
* E: K/ s+ c! B/ y - : M: r! y, H6 q- P
- / k/ z: Q, F5 i3 G2 K
- !目标函数;
* K, F% z# e1 e! e% ` - min=@sum(lc(I,J): lucheng(I,J));
. x. T5 Q6 K% I2 `. N- J1 V - 1 C5 k& z! v8 X7 Q) X4 J5 w
- !整数约束;\" |+ P2 p9 R4 [7 R6 q: w2 w
- @for(plan:@bin(a));
0 r: r\" [, Q: J# f - @for(car(I):@free(time(I)) ;0 G2 M8 ?8 _, H- @+ }. [* Z
- 8 c) d# d: E7 S/ x$ c- S& D+ R7 `
- !从零点开始最终回来;* I$ o0 S! ~+ r- W q# ?# [
- @for(car(I):: o a. v\" g7 N4 I6 b( r, T8 M
- a(1,1,I)=1;a(10,10,I)=1);
& F. Z# q9 G$ }5 ?% u - , o9 ]9 Y9 N1 s* G2 Z& y
- !每个任务只能有一个;8 ]* `3 j, d; J( y. J' h% V
- @sum(car(I):, z; |2 B' Z- I/ u$ @& \$ t
- @sum(place(K):6 z- p- v- f$ Z, l; K7 [1 U% }9 }
- @sum(place(J): a(J,K,I))))=1;
2 Z5 O! {6 o: I9 `, \
9 _* ~2 _- E/ v3 G3 [- !每个车的任务最多有四个;
9 t1 Q$ i' S. @\" O9 u - @for(car(I):
: h0 q( h$ `4 p* t% t5 Y& U; Q - @sum(plan(J,K,I):a(J,k,I))<=6);9 u& z4 u% V4 _0 V4 R\" i9 ?
; o9 H8 B' O( M1 S- !前一辆车任务大于等于后一辆车;\" y% e4 n* {3 {* j% F: m% {9 {# K9 w, C
- @for(car(I)|I#lt#1:
# R* w: f, d8 F4 g/ M! X) {- \ - @sum(plan:a(J,K,I))>=@sum(plan:a(J,K,I+1) );9 q9 w# u2 s: M
9 v& G8 d, x' A- ^- !求which 例子 若a(1,:)=[1 0 0 1 0 1 0 0 0 1];
9 ~ W. _& W\" T% }0 [0 v p. \4 t5 M - ! 则 which(1,:)=[1 1 1 4 4 6 6 6 6 10] ;2 n3 I. ~, D. D0 W, Y\" \7 s
- @for(car(I):3 G& L5 b0 [) u! c0 W& Y
- @for(place(kk)|@sum(place(jj):a(kk,jj,I)) #eq#1:/ h- O$ B( w5 M. n7 x' b
- @for(place(jj)|a(kk,jj,I) #eq#1:6 K' t' w$ K- \
- which(I,kk)=jj ););6 X5 a% V6 w3 K7 J2 l! a
- @for(place(kk)|@sum(place(kk):a(kk,jj,I)) #ne#1:
; h4 Y* k& N; N! f$ N - which(I,kk)=which(I,kk-1));* }! s. d( }0 S/ u s9 o& H/ \
- ( N5 c% P: f# t* Z! N& \\" R/ }, Z
- !时间限制;
! J; g$ X# S! K5 @6 m. }- D - @for(car(jjj):
* f7 V0 P1 V1 o# ]8 y6 s( | - time(jjj)>=-10;time(jjj)<=7);( t$ |/ h6 u7 ~: W4 F
- - |- {% ^: }% U6 r\" [
- @for(jiu(jjj):0 v% U) x\" h3 R9 n# \+ ^8 c* t
- ct(I,jjj)=@if(which(I,jjj)#ne#which(I,jjj+1),dt(which(I,jjj),which(I,jjj+1))+timexie(which(I,jjj)),0));* a: W b0 ^+ T. H
- 1 k; E- k3 ~; \* d2 P
- @for(jiu(jjj):% Z- f8 G: I, N) r3 q
- time(I)+ct(I,jjj)>=time1(jjj+1);time(I)+ct(I,jjj)<=time2(jjj+1));8 w, E\" `* u% c
& O+ k. ~+ D, U1 M- b% A( k# }7 k- !求每辆车行驶的距离;* M. z& j' Y U5 D
- @for(jiu(jjj):: \- C7 B4 Y' Z0 \
- lucheng(I,jjj)=@if(which(I,jjj)#ne#which(I,jjj+1),dd(which(I,jjj),which(I,jjj+1)),0) );. x$ t! T1 }; [$ ~
- );););
/ i V& z5 z6 E( W - end
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