练习2-展厅监控问题

   for e=0:1;                    for f=0:1;                         for g=0:1        for h=0:1;                                 for i=0:1;                                     for j=0:1;    for k=0:1;                                             for l=0:1;                                                 for m=0:1; %%进行约束条件判断,将可行解先挑出来%%                                                   if (a+d+f>=1)&(f+h+l>=1)&(a+b+c>=1)&(c+d+e+g>=1)&(g+h+i+j>=1)&(j+m+l>=1)&(b+e+i+k>=1)&(k+m>=1) %%满足条件(即所有展厅均可监控到)的可行解找出来,并求出摄象头数%%                                        plus=sum([a b c d e f g h i j k l m]); %%将所有满足条件的可行解找出,并构成矩阵position,第一列为相应的摄象头数%%                                                       position=[position;plus a b c d e f g h i j k l m];                                                     end