📄 surveyfuhe.m
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function cpp=surveyfuhe(be,en,sd,gd,s,ed)
cpp=[];
been=[be en];
n=0;%环的序号
if ed<=1
return;
end
xy=0;%最小独立附合环的个数
for i=1:ed-1
[i0,j0]=find(been==i);
if isempty(i0)~=1
xy=xy+1;
for j=i+1:ed
fl=0;%从j开始找,当找到i时其值为1
nu=1;
course=zeros(1,gd);
temp1=[];
[i1,j1]=find(been==j);
if isempty(i1)~=1
tem=been(i1,:);
tem=reshape(tem,1,prod(size(tem)));
tem((tem==j))=[];
temp=[j tem];
course(i1)=j;
while 1%正向搜索==================================================================================================================
nu=nu+1;%每循环一次搜索下一个层次,记录符合环点的个数
for ii=tem
if ii==i
m=i;
t=i1(j);
di=s(t);
fl=fl+1;
break;
end
[i2 j2]=find(ii==been);
beentem=been(i2,:);
for jj=1:size(beentem,1);
xxx=beentem(jj,:);
newpoint=xxx(xxx~=ii);
if any(temp==newpoint)==0;
temp=[temp newpoint];
temp1=[temp1 newpoint];
course(i2(jj))=ii;
end
if newpoint==i
fl=fl+1;
m=ii;
t=i2(jj);
di=s(t);
break;
end
end
if fl==1
break;
end
end
tem=temp1;
temp1=[];
if fl==1
break;
end
end
nu=nu+1;
st=[i m];
while 3%反向搜索==================================================================================================
if length(st)==nu;
break;
end
[i1,j1]=find(been==m);
tem=i1';
f2=0;
for k=tem
f2=f2+1;
if course(k)~=0&course(k)~=m
m=course(k);
st=[st m];
di=di+s(k);
break;
end
end
end
n=n+1;
site1{n}=st;
dis(n)=di;
number(n)=nu;
end
end
end
end
%找出xy条最小独立附合路线=============================================================
n=1;
temp=[];st=[];
tem=find(number==min(number));
[i,j]=min(dis(tem));
site{n}=site1{tem(j)};
dist(n)=dis(tem(j));
st=site{n};
temp=[st(1) st(number(tem(j)))];
number(tem(j))=[];
site1(tem(j))=[];
dis(tem(j))=[];
st=[];
for kk=1:size(site1,2)
tem=find(number==min(number));
[i,j]=min(dis(tem));
st=site1{tem(j)};
if (any(temp==st(1))==0)|(any(temp==st(number(tem(j))))==0)
n=n+1;
site{n}=site1{tem(j)};
dist(n)=dis(tem(j));
if any(temp==st(1))==0
temp=[temp st(1)];
else any(temp==st(number(tem(j))))==0
temp=[temp st(number(tem(j)))];
end
end
number(tem(j))=[];
site1(tem(j))=[];
dis(tem(j))=[];
st=[];
end
cpp=site;
return;
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