这是模拟退火的程序,第三行错误,求大神解决
clc,clear
sj0=load('sj.txt');%加载数据
x=sj0(:,1:2:8);x=x(:);
y=sj0(:,2:2:8);y=y(:);
sj=[x y]; d1=[70,40];
sj=[d1;sj;d1];sj=sj*pi/180;%角度化成弧度
d=zeros(102);%距离矩阵d初始化
for i=1:101
for j=i+1:102
d(i,j)=6370*acos(cos(sj(i,1)-sj(j,1))*cos(sj(i,2))*cos(sj(j,2))+sin(sj(i,2))*sin(sj(j,2)));
endendd=d+d';path=[]; long=inf;%巡航路径及长度初始化
rand('state',sum(clock)); %初始化随机数发生器
for j=1:1000
path0=[1 1+randperm(100),102];
temp=0;
for i=1:101
temp=temp+d(path0(i),path0(i+1));
end
if temp<long
path=path0;long=temp;
endende=0,1^30;
L=2000;at=0,999;T-1;
for k=1:L %退火过程
c=2+floor(100*rand(1,2)); %产生新解
c=sort(c);c1=c(1);c2=c(2); %计算代价函数值的增量
df=d(path(c1-1),path(c2))+d(path(c1),path(c2+1))-d(path(c1-1),path(c1))-d(path(c2),path(c2+1));
if df<0 %接受准则
path=[path(1:c1-1),path(c2:-1:c1),path(c2+1:102)];long=long+df;
elseif exp(-df/T)>=rang
path=[path(1:c1-1),path(c2:-1:c1),path(c2+1:102)];long=long+df;
end
T=T*at;
if T<e
break;
endendpath,long %输出巡航路径及路径长度
xx=sj(path,1);yy=sj(path,2);plot(xx,yy,'-*') %画出巡航路径
clc,clear
sj0=load('sj.txt');%加载数据
x=sj0(:,1:2:8);x=x(:);
y=sj0(:,2:2:8);y=y(:);
sj=[x y]; d1=[70,40];
sj=[d1;sj;d1];sj=sj*pi/180;%角度化成弧度
d=zeros(102);%距离矩阵d初始化
for i=1:101
for j=i+1:102
d(i,j)=6370*acos(cos(sj(i,1)-sj(j,1))*cos(sj(i,2))*cos(sj(j,2))+sin(sj(i,2))*sin(sj(j,2)));
endendd=d+d';path=[]; long=inf;%巡航路径及长度初始化
rand('state',sum(clock)); %初始化随机数发生器
for j=1:1000
path0=[1 1+randperm(100),102];
temp=0;
for i=1:101
temp=temp+d(path0(i),path0(i+1));
end
if temp<long
path=path0;long=temp;
endende=0,1^30;
L=2000;at=0,999;T-1;
for k=1:L %退火过程
c=2+floor(100*rand(1,2)); %产生新解
c=sort(c);c1=c(1);c2=c(2); %计算代价函数值的增量
df=d(path(c1-1),path(c2))+d(path(c1),path(c2+1))-d(path(c1-1),path(c1))-d(path(c2),path(c2+1));
if df<0 %接受准则
path=[path(1:c1-1),path(c2:-1:c1),path(c2+1:102)];long=long+df;
elseif exp(-df/T)>=rang
path=[path(1:c1-1),path(c2:-1:c1),path(c2+1:102)];long=long+df;
end
T=T*at;
if T<e
break;
endendpath,long %输出巡航路径及路径长度
xx=sj(path,1);yy=sj(path,2);plot(xx,yy,'-*') %画出巡航路径