共同讨论一个VRP程序,大家一块来,一起用!

tym103

<p><br/>% <br/>% the procedure of ant colony algorithm for VRP <br/>% <br/>% % % % % % % % % % % </p><p>%initialize the parameters of ant colony algorithms <br/>load data.txt; <br/>d=data(:,1:2); <br/>g=data(:,2); </p><p>m=32; % 蚂蚁数 <br/>alpha=1; <br/>belta=4;% 决定tao和miu重要性的参数 <br/>lmda=0; <br/>rou=0.9; %衰减系数 <br/>q0=0.95; <br/>% 概率 <br/>tao0=1/(32*841.04);%初始信息素 <br/>Q=1;% 蚂蚁循环一周所释放的信息素 <br/>defined_phrm=15.0; % initial pheromone level value <br/>QV=100; % 车辆容量 <br/>vehicle_best=round(sum(g)/QV)+1; %所完成任务所需的最少车数 <br/>V=40; </p><p>% 计算两点的距离 <br/>for i=1:32; <br/>for j=1:32; <br/>dist(i,j)=sqrt((d(i,1)-d(j,1))^2+(d(i,2)-d(j,2))^2); <br/>end; <br/>end; </p><p>%给tao miu赋初值 <br/>for i=1:32; <br/>for j=1:32; <br/>if i~=j; <br/>%s(i,j)=dist(i,1)+dist(1,j)-dist(i,j); <br/>tao(i,j)=defined_phrm; <br/>miu(i,j)=1/dist(i,j); <br/>end; <br/>end; <br/>end; </p><p>for k=1:32; <br/>for k=1:32; <br/>deltao(i,j)=0; <br/>end; <br/>end; </p><p>best_cost=10000; <br/>for n_gen=1:50; </p><p><br/>print_head(n_gen); </p><p>for i=1:m; <br/>%best_solution=[]; <br/>print_head2(i); <br/>sumload=0; <br/>cur_pos(i)=1; <br/>rn=randperm(32); <br/>n=1; <br/>nn=1; <br/>part_sol(nn)=1; <br/>%cost(n_gen,i)=0.0; <br/>n_sol=0; % 由蚂蚁产生的路径数量 <br/>M_vehicle=500; <br/>t=0; %最佳路径数组的元素数为0 </p><p>while sumload&lt;=QV; </p><p>for k=1:length(rn); <br/>if sumload+g(rn(k))&lt;=QV; <br/>gama(cur_pos(i),rn(k))=(sumload+g(rn(k)))/QV; <br/>A(n)=rn(k); <br/>n=n+1; <br/>end; <br/>end; </p><p>fid=fopen('out_customer.txt','a+'); <br/>fprintf(fid,'%s %i\t','the current position is:',cur_pos(i)); <br/>fprintf(fid,'\n%s','the possible customer set is:') <br/>fprintf(fid,'\t%i\n',A); <br/>fprintf(fid,'------------------------------\n'); <br/>fclose(fid); </p><p>p=compute_prob(A,cur_pos(i),tao,miu,alpha,belta,gama,lmda,i); <br/>maxp=1e-8; <br/>na=length(A); <br/>for j=1:na; <br/>if p(j)&gt;maxp <br/>maxp=p(j); <br/>index_max=j; <br/>end; <br/>end; </p><p>old_pos=cur_pos(i); <br/>if rand(1)&lt;q0 <br/>cur_pos(i)=A(index_max); <br/>else <br/>krnd=randperm(na); <br/>cur_pos(i)=A(krnd(1)); <br/>bbb=[old_pos cur_pos(i)]; <br/>ccc=[1 1]; <br/>if bbb==ccc; <br/>cur_pos(i)=A(krnd(2)); <br/>end; <br/>end; </p><p>tao(old_pos,cur_pos(i))=taolocalupdate(tao(old_pos,cur_pos(i)),rou,tao0);%对所经弧进行局部更新 </p><p>sumload=sumload+g(cur_pos(i)); </p><p>nn=nn+1; <br/>part_sol(nn)=cur_pos(i); <br/>temp_load=sumload; </p><p>if cur_pos(i)~=1; <br/>rn=setdiff(rn,cur_pos(i)); <br/>n=1; <br/>A=[]; <br/>end; </p><p>if cur_pos(i)==1; % 如果当前点为车场,将当前路径中的已访问用户去掉后,开始产生新路径 <br/>if setdiff(part_sol,1)~=[]; <br/>n_sol=n_sol+1; % 表示产生的路径数,n_sol=1,2,3,..5,6...,超过5条对其费用加上车辆的派遣费用 <br/>fid=fopen('out_solution.txt','a+'); <br/>fprintf(fid,'%s%i%s','NO.',n_sol,'条路径是:'); <br/>fprintf(fid,'%i ',part_sol); <br/>fprintf(fid,'\n'); <br/>fprintf(fid,'%s','当前的用户需求量是:'); <br/>fprintf(fid,'%i\n',temp_load); <br/>fprintf(fid,'------------------------------\n'); <br/>fclose(fid); </p><p>% 对所得路径进行路径内3-opt优化 <br/>final_sol=exchange(part_sol); </p><p>for nt=1:length(final_sol); % 将所有产生的路径传给一个数组 <br/>temp(t+nt)=final_sol(nt); <br/>end; <br/>t=t+length(final_sol)-1; </p><p>sumload=0; <br/>final_sol=setdiff(final_sol,1); <br/>rn=setdiff(rn,final_sol); <br/>part_sol=[]; <br/>final_sol=[]; <br/>nn=1; <br/>part_sol(nn)=cur_pos(i); <br/>A=[]; <br/>n=1; </p><p>end; <br/>end; </p><p>if setdiff(rn,1)==[];% 产生最后一条终点不为1的路径 <br/>n_sol=n_sol+1; <br/>nl=length(part_sol); <br/>part_sol(nl+1)=1;%将路径的最后1位补1 </p><p>% 对所得路径进行路径内3-opt优化 <br/>final_sol=exchange(part_sol); </p><p>for nt=1:length(final_sol); % 将所有产生的路径传给一个数组 <br/>temp(t+nt)=final_sol(nt); <br/>end; </p><p>cost(n_gen,i)=cost_sol(temp,dist)+M_vehicle*(n_sol-vehicle_best); %计算由蚂蚁i产生的路径总长度 </p><p>for ki=1:length(temp)-1; <br/>deltao(temp(ki),temp(ki+1))=deltao(temp(ki),temp(ki+1))+Q/cost(n_gen,i); <br/>end; </p><p>if cost(n_gen,i)&lt;best_cost; <br/>best_cost=cost(n_gen,i); <br/>old_cost=best_cost; <br/>best_gen=n_gen; % 产生最小费用的代数 <br/>best_ant=i; %产生最小费用的蚂蚁 <br/>best_solution=temp; <br/>end; </p><p>if i==m; %如果所有蚂蚁均完成一次循环，，则用最佳费用所对应的路径对弧进行整体更新 <br/>for ii=1:32; <br/>for jj=1:32; <br/>tao(ii,jj)=(1-rou)*tao(ii,jj); <br/>end; <br/>end; </p><p>for kk=1:length(best_solution)-1; <br/>tao(best_solution(kk),best_solution(kk+1))=tao(best_solution(kk),best_solution(kk+1))+deltao(best_solution(kk),best_solution(kk+1)); <br/>end; <br/>end; </p><p>fid=fopen('out_solution.txt','a+'); <br/>fprintf(fid,'%s%i%s','NO.',n_sol,'路径是:'); <br/>fprintf(fid,'%i ',part_sol); <br/>fprintf(fid,'\n'); <br/>fprintf(fid,'%s %i\n','当前的用户需求量是:',temp_load); <br/>fprintf(fid,'%s %f\n','总费用是:',cost(n_gen,i)); <br/>fprintf(fid,'------------------------------\n'); <br/>fprintf(fid,'%s\n','最终路径是:'); <br/>fprintf(fid,'%i-',temp); <br/>fprintf(fid,'\n'); <br/>fclose(fid); <br/>temp=[]; <br/>break; <br/>end; <br/>end; </p><p>end; <br/>end;</p>

aopao

很强大，但是我看不太懂~~
学习中

dongdong111

给个乱码有毛用啊

tracymath

有道理～～