bokner · December 20, 2024 02:40 · bokner · Dec 20, 2024 · bokner · Dec 20, 2024
diff --git a/advent_or_day16.mzn b/advent_or_day16.mzn
 %% Method: we build a directed graph, with edges 
 %% {from, to} representing a pair of consequent non-overlapping tasks. 
 %% That is, `end_time(from) < start_time(to)`  
 %% Then we constrain a list of tasks for each crew as a path in this graph.
 %% Finally, we constrain the lists to be disjoint.

 include "globals.mzn";

 int: T;
 int: C;
 int: N;
 int: E;

 %% Tasks
 set of int: TASKS = 1..T;
 set of int: TASKS0 = 0..T;
 %% Crews
 set of int: CREWS = 1..C;

 %% Pairs of non-overlapping tasks as edges
 set of int: EDGES = 1..E;

 array[EDGES] of int: from;
 array[EDGES] of int: to;
 array[EDGES] of int: weight;

 array[CREWS, EDGES] of var bool: schedule_assignment;
 array[CREWS, TASKS] of var bool: task_assignment;
 array[CREWS] of var TASKS: first_assignment;
 array[CREWS] of var TASKS: last_assignment;

 array[CREWS] of var 0..sum(weight): crew_cost;

 constraint forall(c in CREWS)(
    bounded_dpath(N,
      E,
      from,
      to,
      weight,
      first_assignment[c],
      last_assignment[c],
      row(task_assignment, c),
      row(schedule_assignment, c),
      crew_cost[c])
 );

 %% One crew per task
 constraint forall(t in TASKS)(
  sum(col(task_assignment, t)) = 1
 );

 solve minimize(sum(crew_cost));

 %% Optional: output

 array[CREWS] of var set of TASKS: crew_tasks;
 constraint forall(c in CREWS)(
  link_set_to_booleans(crew_tasks[c], row(task_assignment, c))
 );
 output ["crew: \(c), task(s): \(crew_tasks[c])\n" | c in CREWS] ++ ["\ntotal cost: \(_objective)"];
	%% Method: we build a directed graph, with edges
	%% {from, to} representing a pair of consequent non-overlapping tasks.
	%% That is, `end_time(from) < start_time(to)`
	%% Then we constrain a list of tasks for each crew as a path in this graph.
	%% Finally, we constrain the lists to be disjoint.

	include "globals.mzn";

	int: T;
	int: C;
	int: N;
	int: E;

	%% Tasks
	set of int: TASKS = 1..T;
	set of int: TASKS0 = 0..T;
	%% Crews
	set of int: CREWS = 1..C;

	%% Pairs of non-overlapping tasks as edges
	set of int: EDGES = 1..E;

	array[EDGES] of int: from;
	array[EDGES] of int: to;
	array[EDGES] of int: weight;

	array[CREWS, EDGES] of var bool: schedule_assignment;
	array[CREWS, TASKS] of var bool: task_assignment;
	array[CREWS] of var TASKS: first_assignment;
	array[CREWS] of var TASKS: last_assignment;

	array[CREWS] of var 0..sum(weight): crew_cost;

	constraint forall(c in CREWS)(
	bounded_dpath(N,
	E,
	from,
	to,
	weight,
	first_assignment[c],
	last_assignment[c],
	row(task_assignment, c),
	row(schedule_assignment, c),
	crew_cost[c])
	);

	%% One crew per task
	constraint forall(t in TASKS)(
	sum(col(task_assignment, t)) = 1
	);

	solve minimize(sum(crew_cost));

	%% Optional: output

	array[CREWS] of var set of TASKS: crew_tasks;
	constraint forall(c in CREWS)(
	link_set_to_booleans(crew_tasks[c], row(task_assignment, c))
	);
	output ["crew: \(c), task(s): \(crew_tasks[c])\n" \| c in CREWS] ++ ["\ntotal cost: \(_objective)"];