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

if n(xb)~=2 if 1

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

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

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(newm) zhuang(qa+1)=zh5 end end end

if sum(n)<(m-1) for i=2:m

[qqqa,qqqb]=size(solut)

kk=find(i==solut) if isempty(kk)

solut(qqqa+1,1)=i n(qqqa+1)=1

zhuang(qqqa+1)=A(3,i) end end end

[op,ok]=size(solut) for i=1:op

solut(i,n(i)+1)=1 end

solut=[ones(op,1) solut]

opt=0;

for i=1:op

for j=2:(n(i)+2) med=c(solut(i,j-1),solut(i,j)) opt=opt+med end end

solut=solut-1

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