节约里程法matlab
clear
clc
A [0 15 5 10 5 20;0 30 30 20 5 10;0 85 162 26 140 110];
rong 300;
m 6;
c zeros 6,6 ;
for i 1:m for j 1:m c i,j sqrt A 1,i -A 1,j ^2+ A 2,i -A 2,j ^2 ; end
end
p zeros 6,6 ;
for i 2: m-1 for j i+1 :m p i,j c 1,i +c 1,j -c i,j ; end
end
s p : ; [hs,wz] sort s,1,'descend' ;
for i 1: m^2 [x,y] ind2sub size p ,wz i ; if A 3,x +A 3,y rong solut [x,y]; n 1 2; zhuang 1 A 3,x +A 3,y ; ii i; break else continue end
end for rr ii+1 : m^2 [x,y] ind2sub size p ,wz rr if hs rr 0 break end [xa,ya] find x solut [xb,yb] find y solut [sa,sb] size solut if isempty xa 0 && isempty xb 0 if xa xb continue end if n xa ~ 2 if 1 ya n xa continue end end if n xb ~ 2 if 1 yb n xb continue end end zh1 zhuang xa zh2 zhuang xb if zhuang xa +zhuang xb rong if ya 1&&yb 1 newm [solut xb,n xb :-1:1 solut xa,1:n xa ] elseif ya 1&&yb n xb newm [solut xb,1:n xb solut xa,1:n xa ] elseif ya n xa &&yb 1 newm [solut xa,1:n xa solut xb,1:n xb ] elseif ya n xa &&yb n xb newm [solut xa,1:n xa solut xb,n xb :-1:1 ] end solut xa,: [] solut xb,: [] n xa [] n xb [] zhuang xa [] zhuang xb [] [qa,qb] size solut solut qa+1 ,1:length newm newm n qa+1 length newm zhuang qa+1 zh1+zh2 else continue end elseif isempty xa 0 if n xa ~ 2 if 1 ya n xa continue end end zh3 zhuang xa +A 3,y if zhuang xa +A 3,y rong if ya 1 newm [y solut xa,1:n xa ] elseif ya n xa newm [solut xa,1:n xa y] end solut xa,: [] n xa [] zhuang xa [] [qa,qb] size solut solut qa+1 ,1:length newm newm n qa+1 length newm zhuang qa+1 zh3 end elseif isempty xb 0 if n xb ~ 2 if 1 yb n xb continue end end zh4 zhuang xb +A 3,x if zhuang xb +A 3,x rong if yb 1 newm [x solut xb,1:n xb ] elseif yb n xb newm [solut xb,1:n xb x] end solut xb,: [] n xb [] zhuang xb [] [qa,qb] size solut solut qa+1 ,1:length newm newm n qa+1 length newm zhuang qa+1 zh4 end else zh5 A 3,x +A 3,y if zh5 rong newm [x y] [qa,qb] size solut solut qa+1 ,1:length newm newm n qa+1 length
