我必须扩展项目的输出和解决方案（进行考试安排）：
- 将结构扩展到五天（我总是在一天工作）： 我想到了时间段（5 * 10）的天数，然后我调整输出！还有更好的方法吗？
现在整个代码：
include "globals.mzn";include "alldifferent.mzn";
%------------------------------Scalar_data----------------------
int: Students; % number of students
int: Exams; % number of exams
int: Rooms; % number of rooms
int: Slotstime; % number of slots
int: Days; % a period i.e. five days
int: Exam_max_duration; % the maximum length of any exam (in slots)
%------------------------------Vectors--------------------------
array[1..Rooms] of int : Rooms_capacity;
array[1..Exams] of int : Exams_duration; % the duration of written test
array[1..Slotstime, 1..Rooms] of 0..1: Unavailability;
array[1..Students,1..Exams] of 0..1: Enrollments;
注册记录每位学生的注册情况; 从中我获得了将参加考试的学生人数， 为了根据容量选择合适的房间
%---------------------------Decision_variables------------------
array[1..Slotstime,1..Rooms] of var 0..Exams: Timetable_exams;
array[1..Exams] of var 1..Rooms: ExamsRoom;
array[1..Exams] of var 1..Slotstime: ExamsStart;
%---------------------------Constraints--------------------------
% Calculate the number of subscribers and assign classroom
% according to time and capacity
constraint forall (e in 1..Exams,r in 1..Rooms,s in 1..Slotstime)
(if Rooms_capacity[r] <= sum([bool2int(Enrollments[st,e]>0)| st in 1..Students])
then Timetable_exams[s,r] != e
else true
endif
);
% Unavailability OK
constraint forall(c in 1..Slotstime, p in 1..Rooms)
(if Unavailability[c,p] == 1
then Timetable_exams[c,p] = 0
else true
endif
);
% Assignment exams according with rooms and slotstimes (Thanks Hakan)
constraint forall(e in 1..Exams) % for each exam
(exists(r in 1..Rooms) % find a room
( ExamsRoom[e] = r /\ % assign the room to the exam
forall(t in 0..Exams_duration[e]-1)
% assign the exam to the slotstimes and room in the timetable
(Timetable_exams[t+ExamsStart[e],r] = e)
)
)
/\ % ensure that we have the correct number of exam slots
sum(Exams_duration) = sum([bool2int(Timetable_exams[t,r]>0) | t in 1..Slotstime,
r in 1..Rooms]);
%---------------------------Solver--------------------------
solve satisfy;
% solve::int_search([Timetable_exams[s, a] | s in 1..Slotstime, a in
% 1..Rooms],first_fail,indomain_min,complete) satisfy;
现在输出，非常沉重和充满弦乐。
%---------------------------Output--------------------------
output ["\n" ++ "MiniZinc paper: Exams schedule " ++ "\n" ]
++["\nDay I \n"]++
[
if r=1 then "\n" else " " endif ++
show(Timetable_exams[t,r])
| t in 1..Slotstime div Days, r in 1..Rooms
]
++["\n\nDay II \n"]++
[
if r=1 then "\n" else " " endif ++
show(Timetable_exams[t,r])
| t in 11..((Slotstime div Days)*2), r in 1..Rooms
]
++["\n\nDay III \n"]++
[
if r=1 then "\n" else " " endif ++
show(Timetable_exams[t,r])
| t in 21..((Slotstime div Days)*3), r in 1..Rooms
]
++["\n\nDay IV \n"]++
[
if r=1 then "\n" else " " endif ++
show(Timetable_exams[t,r])
| t in 31..((Slotstime div Days)*4), r in 1..Rooms
]
++["\n\nDay V \n"]++
[
if r=1 then "\n" else " " endif ++
show(Timetable_exams[t,r])
| t in 41..Slotstime, r in 1..Rooms
]
++[ "\n"]++
[
"\nExams_Room: ", show(ExamsRoom), "\n",
"Exams_Start: ", show(ExamsStart), "\n",
]
++["Participants: "]++
[
if e=Exams then " " else " " endif ++
show (sum([bool2int(Enrollments[st,e]>0)| st in 1..Students]))
|e in 1..Exams
];
我完成了数据：
%Data
Slotstime=10*Days;
Students=50;
Days=5;
% Exams
Exams = 5;
Exam_max_duration=4;
Exams_duration = [4,1,2,3,2];
% Rooms
Rooms = 4;
Rooms_capacity = [20,30,40,50];
Unavailability = [|0,0,0,0 % Rooms rows % Slotstime columns
|0,0,0,0
|0,0,0,0
|0,0,0,0
|1,1,1,1
|1,1,1,1
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
% End first day
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
|1,1,1,1
|1,1,1,1
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
% End secon day
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
|1,1,1,1
|1,1,1,1
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
% End third day
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
|1,1,1,1
|1,1,1,1
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
% End fourth day
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
|1,1,1,1
|1,1,1,1
|0,0,0,0
|0,0,0,0
|0,0,0,0
|0,0,0,0
%End fifth day
|];
Enrollments= [|1,0,1,0,1 % Exams rows %Students columns
|1,0,1,0,1
|0,1,0,0,0
|1,0,0,1,0
|0,1,0,0,0
|0,0,1,1,0
|1,0,0,1,0
|0,0,0,0,1
|1,0,0,0,1
|0,0,0,0,1
|0,1,0,0,0
|0,0,0,0,0
|0,1,0,0,1
|0,0,1,0,1
|1,0,1,0,1
|1,0,1,0,1
|0,1,0,0,0
|1,0,0,1,0
|0,1,0,0,0
|0,0,1,1,0
|1,0,0,1,0
|0,0,0,0,1
|1,0,0,0,1
|0,0,0,0,1
|0,1,0,0,0
|0,0,0,0,0
|0,1,0,0,1
|0,0,1,0,1
|1,0,1,0,1
|1,0,1,0,1
|0,1,0,0,0
|1,0,0,1,0
|0,1,0,0,0
|0,0,1,1,0
|1,0,0,1,0
|0,0,0,0,1
|1,0,0,0,1
|0,0,0,0,1
|0,1,0,0,0
|0,0,0,0,0
|0,1,0,0,1
|0,0,1,0,1
|1,0,1,0,1
|1,0,1,0,1
|0,1,0,0,0
|1,0,0,1,0
|0,1,0,0,0
|0,0,1,1,0
|1,0,0,1,0
|0,0,0,0,1
|];
提前致谢
答案 0 :(得分：5)
对于输出部分，以下代码应该有效。我只改变了日程安排，其余的没有改变。
output ["\n" ++ "MiniZinc paper: Exams schedule " ++ "\n" ]
++
[
if t mod 10 = 1 /\ r = 1 then
"\n\nDay " ++ show(d) ++ " \n"
else "" endif ++
if r=1 then "\n" else " " endif ++
show(Timetable_exams[t,r])
| d in 1..Days, t in 1+(d-1)*10..(Slotstime div Days)*d, r in 1..Rooms,
]
++[ "\n"]++
[
"\nExams_Room: ", show(ExamsRoom), "\n",
"Exams_Start: ", show(ExamsStart), "\n",
]
++["Participants: "]++
[
if e=Exams then " " else " " endif ++
show (sum([bool2int(Enrollments[st,e]>0)| st in 1..Students]))
|e in 1..Exams
];
如果要求日期应使用＆＃34; I＆＃34;，＆＃34; II＆＃34;等编号，那么您可以使用日期名称定义字符串数组，例如< / p>
array[1..Days] of string: DaysStr = ["I","II","III","IV","V"];
然后在输出循环中使用它：
% ....
if t mod 10 = 1 /\ r = 1 then
"\n\nDay " ++ DaysStr[d] ++ " \n" % <---
else "" endif ++
% ....
稍后更新：
使模型更通用（和更小）的另一个原因是用以下方法替换巨大的不可用性矩阵（以及使用它的约束）：
set of int: UnavailabilitySlots = {5,6};
% ....
constraint
forall(c in 1..Slotstime, p in 1..Rooms) (
if c mod 10 in UnavailabilitySlots then
Timetable_exams[c,p] = 0
else
true
endif
);
又一评论：
原始模型有一个缺陷，即允许两天以上的考试，例如：第一天的最后两小时和第二天的前两小时。我认为以下额外（而不是那么漂亮）约束将解决这个问题。再一次，魔术＆＃34; 10＆＃34;使用。
constraint
% do not pass over a day limit
forall(e in 1..Exams) (
not(exists(t in 1..Exams_duration[e]-1) (
(ExamsStart[e]+t-1) mod 10 > (ExamsStart[e]+t) mod 10
))
)
;