function newpop=crossover(pop,pcro,popsize,citysize)
%交叉函數(shù) 
%pcro 為交叉概率 
 
k=1; 
i=0; 
while (k<=popsize)         %按交叉概率選取兩個(gè)個(gè)體
    rk=rand(); 
    if rk<pcro 
        b(i+1)=k; 
        i=i+1; 
    end 
    k=k+1; 
    if i==2 
       pos1=1;
        pos2=1;
        while pos1==pos2
        pos1=fix(rand()*citysize-1)+1;   %隨機(jī)產(chǎn)生第一個(gè)交叉點(diǎn) 
        pos2=fix(rand()*citysize-1)+1;   %隨機(jī)產(chǎn)生第二個(gè)交叉點(diǎn) 
        end
        if pos1>pos2
            temp=pos1;
            pos1=pos2;
            pos2=temp;
        end
        length=pos1-1+citysize-pos2;
        %X1=zeros(1,citysize-1);           %存儲(chǔ)第二個(gè)交叉?zhèn)€體的OX順序
        %X2=zeros(1,citysize-1);           %存儲(chǔ)第一個(gè)交叉?zhèn)€體的OX順序
        
        X1(1:citysize-pos2)=pop(b(2),pos2+1:citysize);
        X1(citysize-pos2+1:citysize)=pop(b(2),1:pos2);
        X2(1:citysize-pos2)=pop(b(1),pos2+1:citysize);
        X2(citysize-pos2+1:citysize)=pop(b(1),1:pos2);
     %刪除與交叉節(jié)點(diǎn)值相同的值
        for t=pos1:pos2
          for s=1:citysize         
               if X1(s)==pop(b(1),t)
                   for r=s:citysize-1
                       X1(r)=X1(r+1);
                   end
               end
               if X2(s)==pop(b(2),t)
                   for r=s:citysize-1
                       X2(r)=X2(r+1);
                   end
               end
           end
        end
   %對(duì)個(gè)體進(jìn)行OX交叉       
          pop(b(1),1:pos1-1)=X1(1:pos1-1);                 
          pop(b(1),pos2+1:citysize)=X1(pos1:length);
          pop(b(2),1:pos1-1)=X2(1:pos1-1);
          pop(b(2),pos2+1:citysize)=X2(pos1:length);
       i=0; 
    end 
end 
 newpop=pop;