在这一章节，我们将用到minizinc， 类似于一个编译器，但是它提供不同的solver帮助我们解决问题，我们只需要对问题中的限制（binary constraint network）进行描述(1.constant 2.variable 3.constraint)，要求solver 为我们提供满足解或者最优解。
生活中我们会遇到很多资源分配问题，例如
在departure point， 有两种不同类型的车，一种只运常温货物，一种可以运常温和低温货物。每辆车都有各自的capacity
如今送货到很多家商店，每个商店都有对于货物的不同需求。
对于不同的问题条件，请给出不同的解决方案。
1.Given the data on the capacities of trucks, and whether they are refrigerated or not, and given the orders placed by some customers, find an allocation of goods to trucks which satisfies all customers orders without exceeding the capacity of any truck and without allocating chilled goods to any non-refrigerated truck.
Answer:
int: C; % Number of customers
int: T; % Number of trucks
int: G; % Number of goods types
int: MAXCAP; % Upper bound on truck capacity
set of int: trucks = 1..T; % Set of trucks
set of int: customers = 0..C; % Set of customers. Includes depot as customer 0
set of int: goods = 1..G; % Set of goods types
int: chilled = 1; int: ambient = 2; % Good types
array[trucks] of int: cap; % Capacity of trucks
array[trucks] of bool: refrig; % Whether or not trucks are refrigerated
array[goods,customers] of int: order; % Number of units of goods types ordered by customers
% Insert your variables and constraints here
var trucks: truck_id; % id of a specific car
var customers: customer_id; % id of a specific customer
array [ trucks, customers] of var 0..MAXCAP: chilled_amount; % an array of ( truck_id, customer_id) means how many chilled goods a truck carry to a customers.
array [ trucks, customers] of var 0..MAXCAP: ambient_amount; % an array of ( truck_id, customer_id) means how many ambient goods a truck carry to a customers.
constraint forall (truck_id in trucks)(if refrig[truck_id] = true then cap[truck_id] >= sum(customer_id in 1..C)(chilled_amount [truck_id, customer_id] + ambient_amount [truck_id, customer_id] ) else cap[truck_id] >= sum(customer_id in 1..C)(ambient_amount [truck_id, customer_id] + chilled_amount [truck_id, customer_id]) endif); % the truck is not overfilled.
constraint forall (customer_id in customers)(order[ambient, customer_id]= sum(truck_id in 1..T)( ambient_amount[truck_id, customer_id]));% all the ambient orders have been achieved.
constraint forall (customer_id in customers)(order[chilled, customer_id] = sum(truck_id in 1..T)( chilled_amount[truck_id, customer_id])); % all the chilled orders have been achieved.
constraint forall(y in 1..T)(if refrig[y] =false then forall(z in 1..C)(chilled_amount[y, z] = 0 ) else true endif); % all the ambient truckscan only take the ambient goods.
% In question Q1, we are only finding a satisfying solution
solve satisfy;
% Write a Minizinc output item to print the solution in the desired format for Q1
output
[show(T)++" "++show(C)++ "\n" ] ++
[ if fix(ambient_amount[truck_id, customer_id]) != 0 \/ fix(chilled_amount[truck_id, customer_id] )!= 0 then show(truck_id) ++","++
show(customer_id) ++","++
show(chilled_amount[truck_id, customer_id]) ++","++
show(ambient_amount[truck_id, customer_id]) ++"\n" else "" endif |truck_id in 1..T, customer_id in 1..C];
Solution：
nb_trucks, nb_customers (not including the depot)
[truck_id, customer_id, chilled_goods_units_delivered, ambient_goods_units_delivered]
%input:
C = 3;
T = 4;
G = 2;
MAXCAP = 10;
% Details of the truck types
cap = [ 7, 8, 8, 10];
refrig = [ false, false, true, true];
% Orders placed by the customers.
order = array2d(goods,customers,
[| 0, 0, 4, 10,
| 0, 4, 8, 7 |]);
%output：
4,3,,
1,3,0,7
2,2,0,8
3,2,4,0
3,3,4,0
4,1,0,4
4,3,6,0
%input
C = 3;
T = 6;
G = 2;
MAXCAP = 10;
% Details of the truck types
cap = [ 5, 5, 7, 7, 6, 10];
refrig = [ false, false, true, true, false, false];
% Orders placed by the customers.
order = array2d(goods,customers,
[| 0, 0, 4, 10,
| 0, 4, 8, 7 |]);
%output
6,3,,
1,1,0,4
3,3,7,0
4,2,4,0
4,3,3,0
5,3,0,5
6,2,0,8
6,3,0,2
%input
C = 4;
T = 12;
G = 2;
MAXCAP = 6;
% Details of the truck types
cap = [ 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 6 ];
refrig = [ false, false, false, false, true, true, true, false, false, true, true, false ];
% Orders placed by the customers.
order = array2d(goods,customers,
[| 0, 2, 0, 4, 10,
| 0, 3, 2, 4, 7 |]);
%output
12,4,
1,3,0,4
1,4,0,1
2,1,0,3
2,2,0,2
5,1,1,0
5,4,4,0
6,3,2,0
6,4,3,0
7,3,2,0
7,4,3,0
10,1,1,0
12,4,0,6
论述：
Time consumption (FD solver):
For 4-3: 40 msec,
For 6-3: 47 msec,
For 12-4: 65 msec,
For 21-8: Cannot finish in 1 min. if we use Gecode solver, then it can finish in 1s.
In this question, I give 4 limitations.
- the truck is not overfilled.
- all the ambient orders have been achieved.
- all the chilled orders have been achieved.
- all the ambient trucks can only take the ambient goods.