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剩余矩形算法的MATLAB实现河海大学12级物理（1）班段付谋1210020111functionmathmode_lsize=zeros(25,2);%表示小矩形的尺寸.M(i,1)、M(i,2)分别表示序号为i的小矩形的宽和高.D=zeros(25,2);%排样方案数组.D(i,1)、D(i,2)分别表示第i个矩形的序号以及r(i).rest=zeros(50,4);%剩余矩形数组.记录每个矩形的左下角坐标（x,y）、宽和高.A=zeros(25,4);%记录数组（记录每个矩形件在样板上的位置）.%初始值均为零。（zeros）size=[12,6;4,7;6,7;10,2;2,5;6,4;4,2;4,6;7,9;4,5;6,4;4,6;6,3;4,5;2,4;8,4;8,6;8,3;6,3;2,6;8,2;3,5;2,5;3,4;2,4;];%小矩形件的尺寸.P1=zeros(1,25);P2=zeros(1,25);P3=zeros(1,25);R1=zeros(1,25);R2=zeros(1,25);R3=zeros(1,25);P1=[4,2,1,3,6,5,7,9,8,10,11,12,14,13,19,15,18,17,20,16,21,22,24,23,25];R1=[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];P2=[10,5,1,13,23,24,22,8,14,4,7,25,11,19,6,2,16,20,18,9,17,3,12,15,21];R2=[0,1,1,1,0,1,1,0,1,1,1,0,0,1,1,1,0,1,0,0,0,1,1,0,0];P3=[23,21,20,16,17,2,24,25,9,3,5,8,22,14,15,18,7,6,10,19,4,12,11,13,1];R3=[0,0,1,0,1,1,0,0,0,1,0,1,0,0,0,1,0,1,1,0,1,0,1,0,0];D1=[P1',R1'];D2=[P2',R2'];D3=[P3',R3'];%至此，数据初始化完毕D=D3;%选择排样方案.w=zeros(25,1);h=zeros(25,1);%矩形件的宽和高N=1;%N是剩余矩形的个数.rest(1,:)=[0,0,15,60];%样板的尺寸设为：宽15，高60.H=0;%占用高度，用来求样板利用率.fori=1:25%矩形件i的宽和高w(i)=size(D(i,1),1+D(i,2));h(i)=size(D(i,1),2-D(i,2));endre=zeros(30,4);fori=1:25%放置25个矩形件%i=3;n=1;j=0;whilen=N%求能包含小矩形件的剩余矩形re，j表示个数.ifrest(n,3)=w(i)&&rest(n,4)=h(i)j=j+1;re(j,:)=rest(n,:);endn=n+1;endk=2;whilek=j%求用到的剩余矩形（放在re的第一行），根据BL条件.ifre(k,2)re(1,2)||(re(k,2)==re(1,2)&&re(k,1)re(1,1))re(1,:)=re(k,:);endk=k+1;endA(i,1)=re(1,1);A(i,2)=re(1,2);A(i,3)=w(i);A(i,4)=h(i);%第i个矩形件的位置信息ifH(A(i,2)+A(i,4))%占用高度的更新.H=(A(i,2)+A(i,4));end%剩余矩形数组的处理更新.NA(i,)n=1;Now=N;whilen=Now;m=0;k=N+1;%if(A(i,2)rest(n,2))&&(A(i,2)(rest(n,2)+rest(n,4)))&&(A(i,1)rest(n,1)+rest(n,3))&&(A(i,1)+A(i,3)rest(n,1))N=N+1;rest(k,1)=rest(n,1);rest(k,2)=rest(n,2);rest(k,3)=rest(n,3);rest(k,4)=A(i,2)-rest(n,2);k=k+1;m=1;endifrest(n,1)+rest(n,3)A(i,1)+A(i,3)&&A(i,1)+A(i,3)rest(n,1)&&(A(i,2)rest(n,2)+rest(n,4))&&(A(i,2)+A(i,4)rest(n,2))N=N+1;rest(k,1)=A(i,1)+A(i,3);rest(k,2)=rest(n,2);rest(k,3)=rest(n,1)+rest(n,3)-A(i,1)-A(i,3);rest(k,4)=rest(n,4);k=k+1;m=1;endifA(i,1)rest(n,1)&&(A(i,1)rest(n,1)+rest(n,3))&&(A(i,2)rest(n,2)+rest(n,4))&&(A(i,2)+A(i,4)rest(n,2))N=N+1;rest(k,1)=rest(n,1);rest(k,2)=rest(n,2);rest(k,4)=rest(n,4);rest(k,3)=A(i,1)-rest(n,1);k=k+1;m=1;endifrest(n,2)+rest(n,4)A(i,2)+A(i,4)&&A(i,2)+A(i,4)rest(n,2)&&(A(i,1)rest(n,1)+rest(n,3))&&(A(i,1)+A(i,3)rest(n,1))N=N+1;rest(k,2)=A(i,2)+A(i,4);rest(k,1)=rest(n,1);rest(k,4)=rest(n,2)+rest(n,4)-A(i,2)-A(i,4);rest(k,3)=rest(n,3);k=k+1;m=1;endifm==0n=n+1;elserest(n,:)=[];N=N-1;Now=Now-1;rest(50,:)=0;%删除要注意！endend%求出剩余矩形数组，下面要对其进行整理.n=1;m=0;whilen=Nk=i+1;whilek=25if(rest(n,3)=w(k)&&rest(n,4)=h(k))m=1;break;elsek=k+1;%endendifm==1n=n+1;elseN=N-1;rest(n,:)=[];rest(50,:)=0;%删去面积为零的或已无法排下所剩的任何一个矩形件的剩余矩形endendn=1;whilen=Nk=1;whilekNifrest(n,1)=rest(k,1)&&rest(n,2)=rest(k,2)&&rest(n,1)+rest(n,3)=rest(k,1)+rest(k,3)&&rest(n,2)+rest(n,4)=rest(k,2)+rest(k,4)ifn~=kN=N-1;m=1;rest(n,:)=[];rest(50,:)=0;%删去被包含的剩余矩形形break;elsek=k+1;m=0;endelsek=k+1;m=0;endendifm==0n=n+1;endend%end%下面进行制图rectangle('position',[0,0,15,60]);axisequal;holdon;x=zeros(1,25);y=zeros(1,25);fori=1:25ifA(i,3)0&&A(i,4)0rectangle('position',A(i,:),'facecolor','g');endx(1,i)=A(i,1)+A(i,3)/2;y(1,i)=A(i,2)+A(i,4)/2;endtext(x,y,num2str(D(:,1)));holdoff;%求板材利用率.a='%';fprintf('板材的利用率为：\n%f%s\n',4000/H,a);end以下是运行结果：方案1排样图-30-20-10010203040010203040506042136579810111214131915181720162122242325D1板材利用率：100%方案2排样图-30-20-10010203040010203040506010511323242281447251119621620189173121521D2板材的利用率：97.560976%方案3的排样图-30-20-10010203040010203040506023212016172242593582214151876101941211131