数学建模社区-数学中国
标题:
求助,代码第二段啥意思
[打印本页]
作者:
魏关亭侯
时间:
2012-7-25 10:30
标题:
求助,代码第二段啥意思
背包问题,12件物品质量为2,5,18,3,2,5,10,4,11,7,14,6,价值为5,10,13,4,3,11,13,10,8,16.7.4,包的最大重量是46
2 @. |- _2 U$ h7 y1 |3 I, W
求最优方案
" ?# i0 n* s6 K# E
代码如下,第二段看不懂,从那个=~开始,那符号啥意思
* @0 j$ V4 V6 W6 u; S8 o
a=0.95
) I! c' Y- f4 h- s. [: r
k=[5;10;13;4;3;11;13;10;8;16;7;4];
# i3 K- I! |; Y/ B/ X3 |
k=-k;
! i6 v3 X5 O! V! _
d=[2;5;18;3;2;5;10;4;11;7;14;6];
! _' c2 E* m5 s$ u* \# F
restriction=46;
8 H$ s R. W2 g2 r3 V( R! Q1 R% s
num=12;
0 W( l4 }( u9 q7 P. m3 j& B
sol_new=ones(1,num);
2 o& V% _& d. M
E_current=inf;E_best=inf;
( {3 ~7 X$ ?2 O
sol_current=sol_new; sol_best=sol_new;
* e' t/ k8 f0 f8 E" V2 F7 X
t0=97;tf=3;t=t0;
* N& J) a% v. h1 m0 I' R
p=1;
3 \, [5 U9 A$ V N& N7 y8 y' S
5 n* H" F$ o9 N! [) d i) U
while t>=tf
% V$ c. N( Z7 A. w7 `# j0 f S* x
for r=1:100
$ e3 [, g9 r; ?' B! x
tmp=ceil(rand.*num);
- S6 x0 {8 w- B1 x& ~- u& v" P0 g
sol_new(1,tmp)=~sol_new(1,tmp);
" ^9 ?, z) R k$ g* b
while 1
1 G Y! t+ q: }
q=(sol_new*d<=restriction)
! z9 x3 ~! l4 x- Y3 C8 X4 \
if ~q
/ w* w" L9 f6 y! H
p=~p
6 z" y! o! }# H* V- K( }
tmp=find(sol_new==1);
6 a8 x2 ]( [ `; l; [. D9 J) d+ O
if p
! u& s1 j9 r" x# s5 D* P
sol_new(1,tmp)=0;
5 H( _/ G" o; B" A
else
( q7 f6 z, l7 P. g: i
sol_new(1,tmp(end))=0;
% g0 W- G9 s1 Z- l
end
/ C& v$ q: x- z9 `
else
0 l: |" `$ W: r4 j0 J# U
break
+ y* V8 p7 Z# @6 j# r4 S
end
/ C: X6 K3 z5 V0 @& H T
end
/ ~8 E" Z( O" o
作者:
赵煜星
时间:
2012-7-25 21:01
好啊 我喜欢
作者:
梦天涯M
时间:
2012-8-5 11:54
model:
2 I0 F, ~$ [6 M% t7 @
sets:
; ~3 G& n5 a! Z
beibao/1..12/:zhiliang,jiazhi,x;
2 P" _8 @4 A; {0 V, W
endsets
8 t8 g5 h1 o& L6 H& |: i) g
data:
6 r; N H0 D4 N p* ~* Y( B
zhiliang=2 5 18 3 2 5 10 4 11 7 14 6;
( I3 R$ H4 u1 o/ o7 ]" k. C
jiazhi=5 10 13 4 3 11 13 10 8 16 7 4;
1 ]: J& i+ \( @/ }) O8 t& T
enddata
: C8 d, n0 G1 l: n
max=@sum(beibao(i):jiazhi(i)*x(i));
* u. [% L) C- X7 A
@sum(beibao(i):zhiliang(i)*x(i))<46;
( H4 E0 f1 v/ @ E8 B5 _
@for(beibao(i):
* R( T+ T) O, M& P0 P
@bin(x(i)));
4 H6 d' a$ x. [( X4 I
+ s' {! v* B: a6 A5 t$ H4 y
结果:
1 l3 {; C4 L; Y) d
Global optimal solution found.
: i5 d. t4 {3 V) n
Objective value: 76.00000
9 p9 l5 k: _. M# ~. Z# T
Objective bound: 76.00000
! M. W. }7 U0 f8 V4 Q6 m- y
Infeasibilities: 0.000000
9 ]3 z7 _% i) S
Extended solver steps: 0
' e7 v$ D2 ?4 {1 A \. m4 _
Total solver iterations: 0
5 v s: n7 @: m) f; q: s4 g% i' W
5 n7 u( v/ f9 _3 z' _
Model Class: PILP
( K4 {. Y+ U9 W: ^, V( U9 E6 O
6 @8 R$ ~: w" t( a
Total variables: 12
% k8 _2 e: t+ _. a* @" Z5 Z1 h
Nonlinear variables: 0
: t- Y9 v5 L! H. z& f/ n! I* F
Integer variables: 12
4 t u' P( C7 E X( v
+ a5 ]' Q3 H# g2 B# A5 O: I
Total constraints: 2
$ p5 T- |1 X( `4 v5 E7 d, R5 n* s" t
Nonlinear constraints: 0
) R. ~/ Y2 r2 d( D" x7 [, D
6 `. F3 z7 @; h1 L% |- v
Total nonzeros: 24
: ?1 G' ?+ }* t) A: K6 m! Z
Nonlinear nonzeros: 0
* G( T Q }% L
" X7 @! l' v1 Q' n- {0 t4 X
6 V" ~. v6 d2 X) c
Variable Value Reduced Cost
" \: a( Z" Q& F
ZHILIANG( 1) 2.000000 0.000000
! h: k$ i2 U" D1 w$ k
ZHILIANG( 2) 5.000000 0.000000
2 I6 I1 i8 q9 E. L
ZHILIANG( 3) 18.00000 0.000000
6 J3 c$ \$ ^( {" z5 i% G& l
ZHILIANG( 4) 3.000000 0.000000
: `3 {! g( o# d7 V! ?9 \
ZHILIANG( 5) 2.000000 0.000000
}6 F: C; U: g) [1 x9 a/ I0 f
ZHILIANG( 6) 5.000000 0.000000
+ s" e8 X1 w# O Z/ @5 ~
ZHILIANG( 7) 10.00000 0.000000
/ t' z/ \7 B9 f# Q
ZHILIANG( 8) 4.000000 0.000000
: x+ j& N2 B# r* m
ZHILIANG( 9) 11.00000 0.000000
. Z/ t& l: O# y
ZHILIANG( 10) 7.000000 0.000000
4 K& B5 r! D7 t2 {
ZHILIANG( 11) 14.00000 0.000000
; Z- d2 ?+ Z, V) L
ZHILIANG( 12) 6.000000 0.000000
% c% G* u: r9 G2 w1 H& T
JIAZHI( 1) 5.000000 0.000000
6 I, l2 N/ d8 S$ ^+ s# Y7 j
JIAZHI( 2) 10.00000 0.000000
+ `4 i% J' }6 i
JIAZHI( 3) 13.00000 0.000000
6 `! O: P) r# z3 w
JIAZHI( 4) 4.000000 0.000000
: ^3 A; S; Z9 T1 A" d
JIAZHI( 5) 3.000000 0.000000
; f1 N. x5 A( l1 g/ R/ N& M ]% D5 P
JIAZHI( 6) 11.00000 0.000000
* S/ L1 d( U1 Z5 K# U/ D( P- d
JIAZHI( 7) 13.00000 0.000000
4 J. ^- L5 J8 {0 I5 h u* }- E1 |
JIAZHI( 8) 10.00000 0.000000
' c' f t# a! V+ Z
JIAZHI( 9) 8.000000 0.000000
/ x6 d" P* k Z$ K" D- p6 N
JIAZHI( 10) 16.00000 0.000000
5 J- _! A4 k2 |+ b; w% [
JIAZHI( 11) 7.000000 0.000000
+ J" n6 F5 z" H& K c
JIAZHI( 12) 4.000000 0.000000
; |, ?9 F3 K' t( l6 t% @
X( 1) 1.000000 -5.000000
/ t! m) ]) T9 J
X( 2) 1.000000 -10.00000
! d8 H3 }" i: ~6 O3 n6 m7 g2 e
X( 3) 0.000000 -13.00000
, g1 E6 `( _/ D! `4 _1 @
X( 4) 1.000000 -4.000000
! l; X& }# b& X1 b* o# |, J
X( 5) 1.000000 -3.000000
9 ~1 x& S; E8 W! z2 X! V& R
X( 6) 1.000000 -11.00000
6 C. w7 ~7 T# ]1 D2 `( U( `) g
X( 7) 1.000000 -13.00000
! Y3 C2 x( l- n
X( 8) 1.000000 -10.00000
, X% J) @, C5 |$ q; V8 w
X( 9) 0.000000 -8.000000
) r* k! S- F9 W/ o+ t4 g( D
X( 10) 1.000000 -16.00000
/ }8 y9 e2 x2 u# }
X( 11) 0.000000 -7.000000
* o1 A# I( ]. z! ]' Y _3 C1 G8 M
X( 12) 1.000000 -4.000000
, u4 U( h9 L7 \: U# Y* T0 y: x
4 d1 O1 H+ F$ v5 i- Z
Row Slack or Surplus Dual Price
: J$ v* f$ c0 ~, _3 d( ~
1 76.00000 1.000000
4 q4 c" @" y/ }/ c, S, p! w5 `! W
2 2.000000 0.000000
* M$ U9 q G$ b R( f* K3 F
! b3 ]5 h3 A8 ^- w+ m
作者:
梦天涯M
时间:
2012-8-5 11:56
不好意思,上边写成bin了,应该是gin
' B3 f% ]. M5 a- Q# Q" v# T$ d/ B0 U
正确程序:
$ X8 o4 }: m5 q( V- F$ P+ d" Z5 N
model:
9 x) g6 ^9 z* c
sets:
) P/ G [3 \' ]6 p
beibao/1..12/:zhiliang,jiazhi,x;
! m2 h8 K2 @1 a. y! f
endsets
! X6 H/ D$ Y- \! w3 V, |. y
data:
& t( P8 {8 ~& l. e
zhiliang=2 5 18 3 2 5 10 4 11 7 14 6;
$ V% }3 ?. D( y
jiazhi=5 10 13 4 3 11 13 10 8 16 7 4;
$ j* R5 x& T" Z2 J/ J8 Z
enddata
. j$ m5 l% m# Z
max=@sum(beibao(i):jiazhi(i)*x(i));
8 v/ x% H3 l0 q( X* e7 r& |
@sum(beibao(i):zhiliang(i)*x(i))<46;
. r& W3 C& S2 [. h6 F$ K5 @
@for(beibao(i):
5 k, @5 { P7 R4 n
@gin(x(i)));
作者:
梦天涯M
时间:
2012-8-5 11:58
Global optimal solution found.
$ \7 b7 Z% u( d* N7 S U/ t: B
Objective value: 115.0000
2 \/ [0 V# C7 k. x, M) w: ]5 d
Objective bound: 115.0000
1 c& r) \" D1 z, [1 X9 Y
Infeasibilities: 0.000000
$ Q' L4 t- G3 {( j) v$ a
Extended solver steps: 0
1 F7 J+ Z8 M1 C1 b6 G( A
Total solver iterations: 0
/ D2 h; w8 y% T$ }, s$ P
$ ~8 y) X: d6 c( V
Model Class: PILP
3 F+ F" I8 l" |/ i, g5 I _
R' V' H* g' T+ Z; S
Total variables: 12
* g. U/ @, D( \. o& R9 A% ~
Nonlinear variables: 0
$ ]9 g7 [9 ] T5 _. S# M
Integer variables: 12
; ?, ]! A1 Y, O. ^
7 N ^$ @9 P: J9 K/ s; n$ T. v* {
Total constraints: 2
' n$ c$ }8 X" i4 j6 x o
Nonlinear constraints: 0
2 m6 y* o2 j# B y
) t' A c7 q7 C. o. P
Total nonzeros: 24
. d3 b* N! X# ?
Nonlinear nonzeros: 0
0 V+ W1 `8 L) s: Y6 S
5 X/ h5 W7 W1 [7 v g7 Y
' x: C) }; o7 V5 t$ T4 b
Variable Value Reduced Cost
) J; v ?1 t" Z
ZHILIANG( 1) 2.000000 0.000000
% c, J" ~% k) y
ZHILIANG( 2) 5.000000 0.000000
: I; x4 W* G9 r5 @$ g. `) s" G
ZHILIANG( 3) 18.00000 0.000000
/ R+ X5 _4 S, ~
ZHILIANG( 4) 3.000000 0.000000
9 S4 r: d/ f5 ]8 \
ZHILIANG( 5) 2.000000 0.000000
/ ~$ @3 ^. j T1 W
ZHILIANG( 6) 5.000000 0.000000
: u- X4 U3 U& j
ZHILIANG( 7) 10.00000 0.000000
! o9 i9 t" m( O+ X
ZHILIANG( 8) 4.000000 0.000000
$ s; L g4 R9 J# {
ZHILIANG( 9) 11.00000 0.000000
# @$ b/ E4 ?1 i) x
ZHILIANG( 10) 7.000000 0.000000
7 D9 X8 w3 h! H# G; N
ZHILIANG( 11) 14.00000 0.000000
1 f$ x" R* ?" M6 `, [
ZHILIANG( 12) 6.000000 0.000000
' j7 p% c. @; y) [4 H# P
JIAZHI( 1) 5.000000 0.000000
8 @6 Q9 @4 P; w0 u; x
JIAZHI( 2) 10.00000 0.000000
- ]7 F0 S* l# R# M; h& W
JIAZHI( 3) 13.00000 0.000000
1 w- n: T9 X8 O
JIAZHI( 4) 4.000000 0.000000
3 \* B4 s1 d) c* q
JIAZHI( 5) 3.000000 0.000000
# |" P! ~# \. o. Q
JIAZHI( 6) 11.00000 0.000000
- v! y6 E- j$ v& k' l
JIAZHI( 7) 13.00000 0.000000
5 B3 k" s8 W3 F' O- D
JIAZHI( 8) 10.00000 0.000000
7 b9 }; m# q h% s2 L% p3 K
JIAZHI( 9) 8.000000 0.000000
) D: {$ Y7 W7 z+ ^( ]5 O
JIAZHI( 10) 16.00000 0.000000
. J% N+ l9 L* u2 [
JIAZHI( 11) 7.000000 0.000000
% _/ J9 _* x. z2 X
JIAZHI( 12) 4.000000 0.000000
3 ?) Q3 d% m) b! F
X( 1) 1.000000 -5.000000
1 `+ R% W4 f! g1 T. |
X( 2) 0.000000 -10.00000
% I8 z, ?( d, H( y& m1 @
X( 3) 0.000000 -13.00000
* ?$ k* I, e5 Z% O
X( 4) 0.000000 -4.000000
& V2 Q3 M: X* {
X( 5) 0.000000 -3.000000
; c* q: U/ L: i/ h1 y, ~
X( 6) 0.000000 -11.00000
# y2 G8 i! e/ k2 ^+ f. t5 z, c r
X( 7) 0.000000 -13.00000
* ~# ?% |. d4 ~! H: t- R
X( 8) 11.00000 -10.00000
7 q8 B6 R2 H3 G
X( 9) 0.000000 -8.000000
2 @* E- }1 }! F2 _* v
X( 10) 0.000000 -16.00000
: U# Q) `, x* i) E. \. j; i z
X( 11) 0.000000 -7.000000
0 M/ V( s1 d5 K
X( 12) 0.000000 -4.000000
# F/ M" |1 C3 Q4 o0 @: N
, J5 {8 ]+ c. f/ ^$ w1 k% a; ?
Row Slack or Surplus Dual Price
) ?, T$ u* X7 Y
1 115.0000 1.000000
6 |" q0 R6 J2 D9 h6 k# A4 D: [
2 0.000000 0.000000
; w }5 N* B/ ?! h G
作者:
梦天涯M
时间:
2012-8-5 11:59
第一件物品取1件,第8件物品取11件,最大价值115
欢迎光临 数学建模社区-数学中国 (http://www.madio.net/)
Powered by Discuz! X2.5