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dekker.one
1int: jobs; % 作业的数量
2set of int: JOB = 1..jobs;
3int: tasks; % 每个作业的任务数量
4set of int: TASK = 1..tasks;
5array [JOB,TASK] of int: d; % 任务持续时间
6int: total = sum(i in JOB, j in TASK)(d[i,j]);% 总持续时间
7int: digs = ceil(log(10.0,total)); % 输出的数值
8array [JOB,TASK] of var 0..total: s; % 起始时间
9var 0..total: end; % 总结束时间
10
11% nooverlap
12predicate no_overlap(var int:s1, int:d1, var int:s2, int:d2) =
13 s1 + d1 <= s2 \/ s2 + d2 <= s1;
14
15constraint %% 保证任务按照顺序出现
16 forall(i in JOB) (
17 forall(j in 1..tasks-1)
18 (s[i,j] + d[i,j] <= s[i,j+1]) /\
19 s[i,tasks] + d[i,tasks] <= end
20 );
21
22constraint %% 保证任务之间没有重叠
23 forall(j in TASK) (
24 forall(i,k in JOB where i < k) (
25 no_overlap(s[i,j], d[i,j], s[k,j], d[k,j])
26 )
27 );
28
29solve minimize end;
30
31output ["end = \(end)\n"] ++
32 [ show_int(digs,s[i,j]) ++ " " ++
33 if j == tasks then "\n" else "" endif |
34 i in JOB, j in TASK ];