商人渡河问题lingo求解

ARRLY

之前，见各种软件求解商人渡河问题，都利用状态转移的思想。但我一直觉得用优化的方法同样是可行的，便小试身手，重新拿起lingo书编了一把：

SETS:
br/1..12/:state;
Num/1..20/;
link(br,Num):Smatrix;
endsets
data:
@OLE('F:\matrix')=Smatrix;

enddata
@for(Num(j): @sum(br(i):smatrix(i,j))=6);!两岸总人数为6;

@for(Num(j):
(Smatrix(1,j)+Smatrix(2,j)+Smatrix(3,j))*((Smatrix(1,j)+Smatrix(2,j)+Smatrix(3,j))#lt#(Smatrix(4,j)+Smatrix(5,j)+Smatrix(6,j)))=0;
(Smatrix(7,j)+Smatrix(8,j)+Smatrix(9,j))*((Smatrix(7,j)+Smatrix(8,j)+Smatrix(9,j))#lt#(Smatrix(10,j)+Smatrix(11,j)+Smatrix(12,j)))=0;
!左右两岸若有商人在，则商人数量不少于仆人;);

@for(num(j)|j #lt# 20:
(Smatrix(1,j)+Smatrix(2,j)+Smatrix(3,j)+Smatrix(4,j)+Smatrix(5,j)+Smatrix(6,j))-(Smatrix(1,j+1)+Smatrix(2,j+1)+Smatrix(3,j+1)+Smatrix(4,j+1)+Smatrix(5,j+1)+Smatrix(6,j+1))<=2;
);
!船的容纳量为2;

@for(link(i,j): @bin(smatrix(i,j)));

Smatrix(1,1)+Smatrix(2,1)+Smatrix(3,1)+Smatrix(4,1)+Smatrix(5,1)+Smatrix(6,1)=6;  !初始状态;
Smatrix(1,20)+Smatrix(2,20)+Smatrix(3,20)+Smatrix(4,20)+Smatrix(5,20)+Smatrix(6,20)=0;!末状态;

min= @sum(link(i,j)|i #le# 6:Smatrix(i,j)); !乘船次数最少;
【题目和其他更多细节不解释】

いがらし

复制不了...太痛苦了

いがらし

楼主能不能整个文件啥的上来？...

宇仲

楼主给力，继续加油！