使用Minizinc选择表中匹配的原料

Question

我有一个具有4个特征（名称，价值，成本，人员）的项目表。我想用Minizinc描述一个模型，该模型选择最大化总价值但总预算为255的项目，并且我可以在20个之间选择近9个项目。

我写一个数据文件：data.dzn

%       name       value    cost  people
data = [
      | PRJ0,      6000,    35,    5
      | PRJ1,      4000,    34,    3
      | PRJ2,      1000,    26,    4
      ...
      | PRJ20,     1200,    18,    2    
      |];


PRJ = {PRJ0,...,PRJ20};
FEATURE = {name, value, budget, personnel};
max_budget = 225;
max_prj=9;

所以我的约束是：

choose_project <= 9 /\ budget<=255 s.t solve maximize tot_value;

如何在表格原始项目中选择非任意数量的决策变量（1到9）？到目前为止，这是我的代码：invest.mzn

include "data.dzn";
int: max_budget;  %255
int: max_prj;     %9
enum FEATURE;
enum PRJ;
array[PRJ,FEATURE] of int: data;
constraint ...
...
solve maximize tot_value;

Answer 1

您可以声明布尔变量数组，例如selected_projects，该数组编码是否选择了项目PRJ_k。

然后您可以简单地计算同时选择此数组中的多少个项目。

示例：

enum FEATURE = {name, value, budget, personnel};
enum PRJ = { PRJ0, PRJ1 };

array[PRJ,FEATURE] of int: data =
     [| PRJ0,      6000,    35,    5
      | PRJ1,      4000,    34,    3
      |];

array[PRJ] of var bool: selected_projects;

var int: tot_value;

% The total number of selected projects must be in [1, 9]
constraint let {
        var int: tot_selected = sum(prj in PRJ) ( selected_projects[prj] )
    } in
        1 <= tot_selected /\ tot_selected <= 9;

constraint tot_value = sum(prj in PRJ) ( selected_projects[prj] * data[prj, value] );

% ...
% encoding of budget and personnel constraints
% ...

solve maximize tot_value;