剩余矩形算法MATLAB實現(xiàn)論文

1、-. z剩余矩形算法的MATLAB實現(xiàn)河海大學12級物理1班段付謀 1210020111function mathmode_lsize=zeros(25,2);%表示小矩形的尺寸.M(i,1)、M(i,2)分別表示序號為 i 的小矩形的寬和高.D=zeros(25,2);%排樣方案數(shù)組.D(i,1)、D(i,2)分別表示第i個矩形的序號以及r(i).rest=zeros(50,4);%剩余矩形數(shù)組.記錄每個矩形的左下角坐標*,y、寬和高.A=zeros(25,4);%記錄數(shù)組記錄每個矩形件在樣板上的位置.%初始值均為零。zerossize=12,6;4,7;6,7;10,2;2,5;6,4;4

2、,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,

3、 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=P

4、2,R2;D3=P3,R3;%至此，數(shù)據(jù)初始化完畢D=D3; %選擇排樣方案.w=zeros(25,1);h=zeros(25,1);%矩形件的寬和高N=1; %N是剩余矩形的個數(shù).rest(1,:)=0,0,15,60;%樣板的尺寸設為：寬15，高60.H=0;% 占用高度，用來求樣板利用率.for i=1:25 %矩形件i的寬和高 w(i)=size(D(i,1),1+D(i,2);h(i)=size(D(i,1),2-D(i,2);endre=zeros(30,4);for i=1:25 %放置25個矩形件%i=3; n=1;j=0;while n=w(i)&rest(n,4)=h(i)

5、 j=j+1;re(j,:)=rest(n,:);end n=n+1;end k=2;while k=j %求用到的剩余矩形放在re的第一行，根據(jù)BL條件.if re(k,2)re(1,2)|(re(k,2)=re(1,2)&re(k,1)re(1,1) re(1,:)=re(k,:); end k=k+1;end A(i,1)=re(1,1); A(i,2)=re(1,2);A(i,3)=w(i);A(i,4)=h(i);%第i個矩形件的位置信息if H(A(i,2)+A(i,4) % 占用高度的更新. H=(A(i,2)+A(i,4);end%剩余矩形數(shù)組的處理更新.N A(i,) n=1

6、;Now=N;while nrest(n,2)&(A(i,2)(rest(n,2)+rest(n,4)&(A(i,1)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;endif rest(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) N=N+1; rest(k,1)=A(i,1)+A(i,3);rest(k,2)=rest(

7、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;endif A(i,1)rest(n,1)&(A(i,1)rest(n,1)+rest(n,3)&(A(i,2)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;endif rest(n,2)+rest(n,4)A(i,2)+A(i,4)&A(i,2)+

8、A(i,4)rest(n,2)&(A(i,1)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;endif m=0 n=n+1;else rest(n,:)=;N=N-1;Now=Now-1;rest(50,:)=0;%刪除要注意！endend%求出剩余矩形數(shù)組，下面要對其進展整理. n=1;m=0;while n=N k=i+1;while k=w(k)&rest(n,

9、4)=h(k) m=1;break;else k=k+1;%endendif m=1 n=n+1;else N=N-1;rest(n,:)=;rest(50,:)=0; %刪去面積為零的或已無法排下所剩的任何一個矩形件的剩余矩形endend n=1;while n=N k=1;while k=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)0&A(i,4)0 rectangle(position,A(i,:),facecolor,g); end *(1,i)=A(i,1)+A(i,3)/2;y(1,i)=A(i,2)+A(i,4)/2;end