資源簡介
可以直接運行求解VRP的matlab程序,可以求解具有時間窗、與容量限制的車輛路徑問題
代碼片段和文件信息
function?[]=antvrptw()
clc
clear
tic%開始計時
city_coordinate=[33042312;36391315;41772244;37121399;34881535;33261556;32381229;41961044;4312790;4386170;
?????????????????30071970;25621756;27881491;23811676;1332695;37151678;39182179;40612370;37802212;36762578;
?????????????????30291838];%坐標???????????
nodedemand=[09040607070402040706040404020805030505050];%需求量
timewindow=[0290240260270270240220240270260240340340320380350230250250350];%時間窗限制
servicewindow=[09040607070402040706040404020805030505050];%服務時間
m=30;%?m?螞蟻個數
Alpha=3;%?Alpha?表征信息素重要程度的參數
Beta=2;%?Beta?表征啟發式因子重要程度的參數
gama=2;
Rho=0.2;%?Rho?信息素蒸發系數
NC_max=50;
Q=2;%?Q?信息素增加強度系數
W=300;
qq=0.1;
load_w=0;
s=0;%??????
C=city_coordinate;
demands=nodedemand;
windowtime=timewindow;
servicetime=servicewindow;
vehicletime=0;
n=size(C1);%n??????????????
D=zeros(nn);%D?????????????
for?i=1:n?
???for?j=1:n?
?????if?i~=j?
?????????D(ij)=((C(i1)-C(j1))^2+(C(i2)-C(j2))^2)^0.5;?
?????else?
????????D(ij)=eps;?
?????end?
????D(ji)=D(ij);?
???end?
end?
Eta=1./D;%Eta????????????????
Tau=ones(nn);%Tau???????
Tabu=zeros(mn+32);%???????????
NC=1;%??????
G_best_route=[NC_maxn+32];%???????
G_best_length=inf.*ones(NC_max1);%??????????
length_ave=zeros(NC_max1);%?????????
%%?????????DC?
while?NC<=NC_max%????????????????
????%Tabu(:1)=randint(m1[11]);%如randint(23[1?6]),就是產生一個2*3隨機矩陣,這個矩陣的元素是區間[1?6]的隨機數。
????Tabu(:1)=randi(1m1);
%%????m?????????????????????
for?i=1:m
????visited=Tabu(i:);
????visited=visited(visited>0);
????to_visit=setdiff(1:nvisited);
????c_temp=length(to_visit);
????j=1;
???????while?j<=n
??????????if?~isempty(to_visit)
???%%??????????????????
????for?k=1:length(to_visit)
????????x(k)=(Tau(visited(end)to_visit(k))^Alpha)*(Eta(visited(end)to_visit(k))^Beta);%*(U(visited(end)to_visit(k))^gama);
??????????%?x(k)=(Tau(visited(end)to_visit(k))^Alpha)*(Eta(visited(end)to_visit(k))^Beta)*(U(visited(end)to_visit(k))^gama);
????end???
?????????ww=rand;
????if?ww ?????????Select=find(max(x));
????el
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