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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:
$ }3 U. G6 c- ~ - !n个货车8收点运输问题;\" ~3 U: U# u; c* v( p4 g8 p
- sets:5 i& ]' I4 S+ l) Z0 u+ g L
- place/place1..place10/:time1,time2,timexie;: F6 I9 x5 U* y+ T
- car/car1..car3/:time;9 S* x. w- h0 `+ m. C- U! M$ w
K; A7 ]' R\" y7 [- jiu/1..9/;
8 Q u$ ^; y% x2 ~% b - er/1..2/;
$ k/ \0 ^' K, R a3 w9 n% {
: j0 b: k: S% n$ H, `- lc(car,jiu):lucheng,ct;1 Z# ]/ K( {# m1 Z
- plan(place,place,car):a;. I& r0 |( v* T9 k\" |! G3 O
- distance(place,place):dd,dt,which;
l, a\" _1 Q6 z' ]6 N/ ?6 Z - endsets
/ h% R4 g1 i8 M# F9 G
3 P' v7 O0 ?4 ~6 _- [- 3 @, V/ C& H! Y! ^# x
- !这里是数据;
4 X, g- g% B\" X/ N! q; [0 G1 } - data:
1 ^/ _& w- E) d/ S* B+ g! a/ ? - s=1;! a; p- T1 m- a1 U/ y9 V- o8 Z
- time1=-10 1 4 1 4 3.5 2 5 1.5 1;
: }9 Q8 s) N) \5 v\" v - time2=7 4 6 2 7 5 5 8 4 10 ;7 m7 t1 b7 u4 _& R
- timexie=0 1 2 1 3 2 2.5 3 0.8 0;
- K0 |: @* `3 O4 q
9 M+ q\" n) x$ M x
, Q2 d$ t/ B0 I: h1 m8 Z$ S1 c- dd=
; ]7 P% H0 L: |- a2 T - 0 40 60 75 90 90 100 135 80 0, q. f N2 U! h
- 40 0 65 40 100 50 75 110 100 40) u& s4 I( A- @: u* i' M
- 60 65 0 75 100 100 75 75 75 60
& D$ l+ [7 ?\" e8 [; S - 75 40 75 0 100 50 90 90 125 75
4 |- F8 D% e7 y# J/ t - 90 100 100 100 0 100 75 75 100 90 M, ?' u. x0 T p+ n
- 90 50 100 50 100 0 70 90 75 90
3 j6 V' q* T. N) }/ p% m8 ` - 100 75 75 90 75 70 0 70 100 100
% A9 \2 N2 r7 j T - 135 110 75 90 75 90 70 0 100 1352 B6 O& U$ F) R2 ?; p/ q1 J
- 80 100 75 125 100 75 100 100 0 80
4 g1 r, j) }! \+ A5 E; p5 _ - 0 40 60 75 90 90 100 135 80 0 ;
9 W$ Z: g3 Z* t5 ]1 \ - dt= 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 02 d1 W0 { ~/ f$ M6 z$ C
- 0.8000 0 1.3000 0.8000 2.0000 1.0000 1.5000 2.2000 2.0000 0.8000+ g' E- i/ q: f; M
- 1.2000 1.3000 0 1.5000 2.0000 2.0000 1.5000 1.5000 1.5000 1.2000, F8 F9 \+ r+ F) z
- 1.5000 0.8000 1.5000 0 2.0000 1.0000 1.8000 1.8000 2.5000 1.5000
$ a4 n3 a5 z7 \8 w7 g( y5 G - 1.8000 2.0000 2.0000 2.0000 0 2.0000 1.5000 1.5000 2.0000 1.80006 {; `' N' D* c: h) L/ r
- 1.8000 1.0000 2.0000 1.0000 2.0000 0 1.4000 1.8000 1.5000 1.8000# z k# \, }. C5 k5 X
- 2.0000 1.5000 1.5000 1.8000 1.5000 1.4000 0 1.4000 2.0000 2.00005 ], O( A V6 E' M
- 2.7000 2.2000 1.5000 1.8000 1.5000 1.8000 1.4000 0 2.0000 2.70000 Y7 G7 g) }) F- y7 G0 v1 M3 b
- 1.6000 2.0000 1.5000 2.5000 2.0000 1.5000 2.0000 2.0000 0 1.6000; z: j5 p8 t- m1 p& f4 P6 e3 Q) u
- 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 0;
3 f: x5 X\" S* r3 t5 m. k\" j: T - enddata
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$ X1 b2 W+ W1 w2 b8 d- !目标函数;
8 P* o( `\" h \9 C - min=@sum(lc(I,J): lucheng(I,J));
7 }% S; i6 L3 m/ _% x c2 r7 K' d; A
; r+ v) f6 ^6 E: p, `- !整数约束;
6 @/ d! H! B: H - @for(plan:@bin(a));5 p) z8 G\" G8 M- \/ s
- @for(car(I):@free(time(I)) ;
- |6 n0 D9 P9 L9 {6 F' m
# u9 Q0 W! |0 z+ D- !从零点开始最终回来;, E/ {9 ~ g( @6 m
- @for(car(I):
\" L) X1 `. r/ `$ b: s' V1 m/ n - a(1,1,I)=1;a(10,10,I)=1);# u2 V! ~5 T$ U4 p9 A. F) s6 N\" m
- 2 E+ q: i# ]* Q7 y4 G6 x; ~3 Y# y- j
- !每个任务只能有一个;, N: W' L m) p2 H/ N
- @sum(car(I):0 X: K3 s$ i) L! k/ r- C
- @sum(place(K):% F8 _8 }; J# d. i4 Y# x
- @sum(place(J): a(J,K,I))))=1;
8 E\" z- o/ N) w+ W - ( Y& S1 ?& S$ a* o\" I
- !每个车的任务最多有四个;
; B# E3 W: P# [$ X - @for(car(I):0 d% ^: A( g2 t( |$ l7 h
- @sum(plan(J,K,I):a(J,k,I))<=6);( [0 x1 Y3 g$ H3 Z, l
- 0 Q. D* [) i9 B
- !前一辆车任务大于等于后一辆车;
) e; S& ~; t% F W. g! Y Z - @for(car(I)|I#lt#1:
( a1 N- J/ f# ^3 g8 s; g - @sum(plan:a(J,K,I))>=@sum(plan:a(J,K,I+1) );8 B) k: a9 T( @
- 8 R4 p7 D9 B1 z& Q/ w
- !求which 例子 若a(1,:)=[1 0 0 1 0 1 0 0 0 1];
# x+ W/ l; Y) Z. M* i - ! 则 which(1,:)=[1 1 1 4 4 6 6 6 6 10] ;
& `. n8 T) W0 x$ j/ y2 J - @for(car(I):
1 W9 ~: e9 P4 Z# E - @for(place(kk)|@sum(place(jj):a(kk,jj,I)) #eq#1:3 |( E, a/ ~+ G% R+ q' K
- @for(place(jj)|a(kk,jj,I) #eq#1:
- w Z- @1 J7 E; i1 z, }# I5 W - which(I,kk)=jj ););2 p) c& |7 N( Y4 Y( G4 r2 B
- @for(place(kk)|@sum(place(kk):a(kk,jj,I)) #ne#1:
4 W, t' `! J+ y2 E - which(I,kk)=which(I,kk-1));1 \0 V9 [4 e. k7 T1 ]6 r2 `
- , G$ F& ], x) Y+ p1 L
- !时间限制;
# s9 ^, d$ r4 L' B5 M2 w - @for(car(jjj):7 v' Z+ G; M D/ K$ h
- time(jjj)>=-10;time(jjj)<=7);
8 N* q1 v3 B. C- C2 ]
3 k+ J, H* O: x7 T0 a- @for(jiu(jjj):
5 o\" i\" l/ s/ @7 w& f - 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));
4 @- m5 E8 N: R, r+ \7 H) b
/ C1 B, d1 i2 X1 V; }2 O- @for(jiu(jjj):% L0 J% c, t' s* R# F
- time(I)+ct(I,jjj)>=time1(jjj+1);time(I)+ct(I,jjj)<=time2(jjj+1));
6 G+ R1 ]/ F; t% X# [# K
! a0 S2 y) v; N) s7 Q. u( e\" |/ w- !求每辆车行驶的距离;
4 {% t2 K& s* }! V\" Y( C - @for(jiu(jjj):
* b( a8 E( V ~; K( J* t, e - lucheng(I,jjj)=@if(which(I,jjj)#ne#which(I,jjj+1),dd(which(I,jjj),which(I,jjj+1)),0) );+ a; H8 T0 l* [% ]0 ? ~7 v
- );););- M9 H5 |+ a- k. ? E
- end ?, h( r% V2 e
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