3d装箱考虑多个物体装入一个箱子，且考虑长宽高体积重量的matlab算法程序

3D装箱问题是一个NP难问题，通常使用启发式算法或者元启发式算法来解决。以下是一个基于贪心算法的Matlab程序示例，可以考虑物体的长宽高、体积和重量。

% 物体信息
object_name = {'A', 'B', 'C', 'D', 'E'};
object_width = [2, 3, 4, 2, 3];
object_height = [3, 2, 3, 4, 2];
object_length = [4, 3, 2, 3, 4];
object_volume = object_width .* object_height .* object_length;
object_weight = [10, 5, 8, 12, 6];

% 箱子信息
box_width = 6;
box_height = 5;
box_length = 8;
box_volume = box_width * box_height * box_length;
box_weight_limit = 20;

% 初始化
box_volume_used = 0;
box_weight_used = 0;
box_contents = cell(1, length(object_name));
object_index = 1:length(object_name);

% 按照体积从大到小排序
[~, sorted_index] = sort(object_volume, 'descend');
sorted_object_name = object_name(sorted_index);
sorted_object_width = object_width(sorted_index);
sorted_object_height = object_height(sorted_index);
sorted_object_length = object_length(sorted_index);
sorted_object_volume = object_volume(sorted_index);
sorted_object_weight = object_weight(sorted_index);

% 装箱
for i = 1:length(sorted_index)
    if box_volume_used + sorted_object_volume(i) <= box_volume && box_weight_used + sorted_object_weight(i) <= box_weight_limit
        box_contents{i} = sorted_object_name(i);
        box_volume_used = box_volume_used + sorted_object_volume(i);
        box_weight_used = box_weight_used + sorted_object_weight(i);
    end
end

% 输出结果
fprintf('Box contents:\n');
for i = 1:length(box_contents)
    if ~isempty(box_contents{i})
        fprintf('%s\n', box_contents{i});
    end
end
fprintf('Remaining volume: %d\nRemaining weight: %d\n', box_volume - box_volume_used, box_weight_limit - box_weight_used);

这个程序将物体按照体积从大到小排序，然后从大到小进行装箱，如果装不下了就跳过该物体。程序输出箱子里的物体列表，以及剩余的容积和重量限制。请注意，这个算法只能保证得到一个可行解，而不一定是最优解。