遗传算法一些应用于编程含源代码tsp问题求解

closeall% %salesmen%min_tour每个旅行商最少的城市%pop_size种群%num_iternargs=7; fork=nargin:nargs-1switchcasecase1salesmen=casemin_tour=casepop_size=casenum_iter=5e3;case5show_prog=case%检查输入矩阵[nr,ncsize(dmat);ifnr~=ncnnr1;%%salesmen=pop_size=max(8,8*ceil(pop_size(1)/8));show_prog=logical(show_prog(1));show_res=logical(show_res(1));%num_brks=salesmen-dof=n- %可以的城市addto=ones(1,dof+1);fork=2:num_brkscum_prob=%pop_rte=zeros(pop_size,n); %路径集合的种群pop_brk=zeros(pop_size,num_brks); %断点集合的种群fork=1:pop_sizepop_brk(k,:)=randbreaks();%global_min=Inf; total_dist=zeros(1,pop_size);dist_history=tmp_pop_rte=zeros(8,n); tmp_pop_brk=zeros(8,num_brks);%当前的断点设置new_pop_rte=zeros(pop_size,n);%更新的路径设置new_pop_brkzeros(pop_size,num_brks);%更新的断点设置ifshow_progpfig=figure('Name','MTSPF_GA|CurrentBestforiter=forp=p_rte=pop_rte(p,:);p_brk=pop_brk(p,:);rng=[[1p_brk+1];[p_brkfors=d(sd(sdmat(1,p_rte(rng(s,1)))/35;%添加开始的路径所用时间(的路城

fork=rng(s,1):rng(s,2)-1 ifk<=36d(sd(s d(sd(s d(sd(sdmat(p_rte(rng(s,2)),1)/35%这一块是加的，即加上每一个商人旅行时间不超过24小时的约束条件 ifforiftotal_dist(p)==0;[min_dist,index]=min(total_dist);= ifmin_dist<global_minopt_rte opt_brk rng=[[1opt_brk+1];[opt_brkn]]'; %forp=8:8:pop_sizertes=pop_rte(rand_grou(p-7:p),:);[ignore,idx]=min(dists);best_of_8_rte=rtes(idx,:);best_of_8_brk=brks(idx,:);I=rte_ins_pts(1);J=fork=1:8%产生新的方案tmp_pop_rte(k,:)=best_of_8_rte;switchkcase2%case3%互换操作tmp_pop_rte(k,[IJtmp_pop_rte(k,[JI]);case4%滑动平移操作tmp_pop_rte(k,I:Jtmp_pop_rte(k,[I+1:JI]);case5%更新断点case6%倒置并更新断点tmp_pop_brk(k,:)=randbreaks();case7%tmp_pop_rte(k,[IJ])=tmp_pop_rte(k,[JI]);tmp_pop_brk(k,:)=randbreaks();case8%tmp_pop_rte(k,I:J)=tmp_pop_rte(k,[I+1:JI]);tmp_pop_brk(k,:)=randbreaks();otherwise

new_pop_rte(p-7:p,:)=tmp_pop_rte;pop_rte=new_pop_rte;%rng=[[1opt_brk+1];[opt_brk fors=1:salesmend=d+dmat(1,opt_rte(rng(s,1)))/35;%添加开始的路径所用时间(到断点的路城)fork=rng(s,1):rng(s,2)-1 ifdd dd

dddmat(opt_rte(rng(s,2)),1)/35ifvarargout{1}=opt_rte;varargout{2}=opt_brk;varargout{3}=global_min;varargout{4}=dis_e;%functionbreaks=ifmin_tour1breaks=sort(tmp_brks(1:num