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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!; K1 p/ y( v$ J2 g
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- model:
) b- q\" Z. A* _- f' G# ^ - !n个货车8收点运输问题;) ?* Z\" `# f/ s& V0 U) R2 F% O
- sets:
\" r$ }' i# d0 u; u# | - place/place1..place10/:time1,time2,timexie;
; }* R$ y9 M7 u) |% |7 @ - car/car1..car3/:time;
+ @ z w7 C4 U# M. Z7 q - % G& r6 i# q' p5 D\" k. [4 u
- jiu/1..9/;
% ?( I! U2 u% F5 ~: y+ n' B! I: c6 j3 M - er/1..2/;
6 |2 `\" n: a- L: ~ - ( i& l+ |: D2 ]7 }
- lc(car,jiu):lucheng,ct;, e& o( `$ Q- G @+ D
- plan(place,place,car):a;\" p$ x: X4 \9 b3 L* g\" M
- distance(place,place):dd,dt,which;
0 U7 {+ D, Z5 u - endsets
+ C6 G2 x0 U E1 V: m: |6 I& s0 O$ ]4 h
+ e; C, S0 u1 {3 w' H3 B( N- . p3 O/ z2 b; J, C s% |+ B
- !这里是数据;
8 H* [6 ], P* ]5 @1 X9 K+ Q - data:/ y' ~: V3 a4 k' w. ~
- s=1;
- V0 U! O. a+ b1 j' m- } - time1=-10 1 4 1 4 3.5 2 5 1.5 1; % p% z5 H; e* k* t: G% l8 L7 B
- time2=7 4 6 2 7 5 5 8 4 10 ;
+ @' F# n, ^% b6 S; k% [ O\" k - timexie=0 1 2 1 3 2 2.5 3 0.8 0;- U3 Z% |) a- b$ ]6 I4 z3 r% i8 x5 N% R
& W# q4 J3 Y6 u2 z O$ @6 A- : R0 a t- C0 J; E( I, N
- dd=
9 R' @' _1 d7 h0 G# _5 D. s) a) Q - 0 40 60 75 90 90 100 135 80 0
* z/ g) z- s }) A+ d. I( q - 40 0 65 40 100 50 75 110 100 40
0 l& d6 y0 K+ H - 60 65 0 75 100 100 75 75 75 60
2 ?9 l/ P% M& w% h. r\" B! t - 75 40 75 0 100 50 90 90 125 75: B) f6 W; }) J% W- Z& d+ u z# }: q ?
- 90 100 100 100 0 100 75 75 100 90* T( ]2 v% Z\" V2 R# Z
- 90 50 100 50 100 0 70 90 75 90
1 B! L t1 y9 h( x: q% S( l - 100 75 75 90 75 70 0 70 100 100' T, p# i' |: q# o: s* @( A( y
- 135 110 75 90 75 90 70 0 100 1355 h) \( q! K0 M1 X5 X: s+ v
- 80 100 75 125 100 75 100 100 0 80
, F7 T- o5 d$ k, i5 s6 a6 q8 g/ l6 ` - 0 40 60 75 90 90 100 135 80 0 ;1 }/ H. W5 S0 ~( i- o8 ?: `
- dt= 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 04 b4 h/ R2 i2 h5 R
- 0.8000 0 1.3000 0.8000 2.0000 1.0000 1.5000 2.2000 2.0000 0.8000
{9 Q9 ?5 K3 i3 e: `$ |3 T - 1.2000 1.3000 0 1.5000 2.0000 2.0000 1.5000 1.5000 1.5000 1.2000
$ B$ G4 P7 G1 V8 S% {; U - 1.5000 0.8000 1.5000 0 2.0000 1.0000 1.8000 1.8000 2.5000 1.5000
; ]4 C! e\" D/ Q X7 E; X# N: r; z - 1.8000 2.0000 2.0000 2.0000 0 2.0000 1.5000 1.5000 2.0000 1.8000$ l3 s8 P7 \! H# O% g
- 1.8000 1.0000 2.0000 1.0000 2.0000 0 1.4000 1.8000 1.5000 1.8000
6 ^( T7 X. j( x8 O - 2.0000 1.5000 1.5000 1.8000 1.5000 1.4000 0 1.4000 2.0000 2.0000
) m* X& D# W# E f$ m6 @ - 2.7000 2.2000 1.5000 1.8000 1.5000 1.8000 1.4000 0 2.0000 2.7000
( y+ ?; B4 p2 ~$ k - 1.6000 2.0000 1.5000 2.5000 2.0000 1.5000 2.0000 2.0000 0 1.6000
9 E+ r$ @: y, s: G9 N - 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 0;
. M3 }- Z2 ?9 _6 S T. x: w - enddata
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- !目标函数;
% U4 k( O ~4 f& z1 }5 M! r - min=@sum(lc(I,J): lucheng(I,J));
4 q% v! p0 {6 o# k! i1 l! H+ \ - 5 c7 J1 v' }3 p) [4 n+ ]$ B/ v
- !整数约束;
* V J& ~! ^' D; `8 v( X - @for(plan:@bin(a));% d7 q* d! z8 |+ B5 L- k
- @for(car(I):@free(time(I)) ;
\" S6 a0 ]$ T7 Y; z3 @1 `7 |3 E - * S( x: l7 [0 ?, E\" _$ V8 u' I
- !从零点开始最终回来;. o n- s6 r3 M2 k
- @for(car(I):
3 E$ V2 _- x% A8 m5 ]! d+ M5 I6 T - a(1,1,I)=1;a(10,10,I)=1);
- M* f- z( p) ~- W) Z - . ]1 g: Y1 F' p. G9 \
- !每个任务只能有一个;
7 C- `9 L* W1 j% i; x4 E\" ~ - @sum(car(I):; R7 j4 u8 i3 d
- @sum(place(K):# E8 C6 v0 i' G8 l- z
- @sum(place(J): a(J,K,I))))=1;
- `) S1 t' B/ r7 p2 Y! K - : t4 ^6 G- ^8 i! t! T
- !每个车的任务最多有四个;' S8 V5 q4 C) |- y# ?9 m( ?6 [
- @for(car(I):, ~* X7 V2 S\" q) D6 J
- @sum(plan(J,K,I):a(J,k,I))<=6);3 d' W' E9 f5 K0 y) w
8 r# I( f8 l q/ H- !前一辆车任务大于等于后一辆车;7 E7 y/ Q5 s% T: q5 o; T1 y
- @for(car(I)|I#lt#1:
! y5 `# `, ^0 }4 k - @sum(plan:a(J,K,I))>=@sum(plan:a(J,K,I+1) );$ o; v7 C! K k7 F8 ]% ~1 _
- S& T T4 M& ~5 P/ _
- !求which 例子 若a(1,:)=[1 0 0 1 0 1 0 0 0 1];
( ^/ H3 e' _\" P - ! 则 which(1,:)=[1 1 1 4 4 6 6 6 6 10] ;0 V\" n6 D. Y5 n% ^% X
- @for(car(I):3 O& _9 a- @9 A9 W1 N( |' x
- @for(place(kk)|@sum(place(jj):a(kk,jj,I)) #eq#1:
- ^+ S; P+ X0 _1 a% { - @for(place(jj)|a(kk,jj,I) #eq#1:% M- }6 X0 b( |: T- q
- which(I,kk)=jj ););
) z3 [' Z( X. Z9 @3 ?! e8 p - @for(place(kk)|@sum(place(kk):a(kk,jj,I)) #ne#1:
3 `% X2 W1 c) C - which(I,kk)=which(I,kk-1));
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0 p- R& L g9 Z5 P3 p' J; G- !时间限制;4 @\" Y( | w! B* f0 x, n5 h5 P, o% Z: S
- @for(car(jjj):
, |0 E. t0 W. A7 K\" @' B; b: |! n, A { - time(jjj)>=-10;time(jjj)<=7);
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- @for(jiu(jjj):
7 A- t/ S, F) d1 M6 E - 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));
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- @for(jiu(jjj):' s; p, }7 C* T: j
- time(I)+ct(I,jjj)>=time1(jjj+1);time(I)+ct(I,jjj)<=time2(jjj+1));
& u3 u5 F) H6 p' Y/ l; T - ! v8 c/ a/ ~( Q3 Z! X8 k U+ ^
- !求每辆车行驶的距离;+ ?, `) u7 G1 U
- @for(jiu(jjj):
) t' i5 X/ \7 m0 u - lucheng(I,jjj)=@if(which(I,jjj)#ne#which(I,jjj+1),dd(which(I,jjj),which(I,jjj+1)),0) );
6 S1 S6 x x. o, R - );););8 { e- L\" v; G+ Y1 c
- end2 O$ Z O9 n5 d+ u0 E; O9 ?
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