我现在正在为一个客户解决以下问题,该客户将创建最经济的时间表(使用最少的替代品),因为:
- 替代者应尽可能连续地代替老师工作(*不是一个大问题)
- 潜艇只能工作 6 个时期
到目前为止,我有一个教师类(如下所示)和一个组织者类,它们实际上创建了最佳时间表。现在我只是让程序在网格中循环填充每个替代品。
Teacher[] t= new Teacher[14];
Organizer o = new Organizer(t);
o.sort();
int[][] g = o.getGrid();
示例输入:
t[0] = new Teacher("Teacher 1", "Mr", new int[]{1,0,1,0,0,0,0});
t[1] = new Teacher("Teacher 2","Mr", new int[]{1,1,0,1,1,0,1});
t[2] = new Teacher("Teacher 3","Mr", new int[]{0,1,1,1,1,1,0});
t[3] = new Teacher("Teacher 4","Mr", new int[]{1,1,0,1,1,0,1});
t[4] = new Teacher("Teacher 5","Mr", new int[]{1,1,0,0,1,1,1});
t[5] = new Teacher("Teacher 6", "Mr", new int[]{1,1,1,0,0,1,1});
t[6] = new Teacher("Teacher 7", "Mr", new int[]{0,0,1,0,1,1,1});
t[7] = new Teacher("Teacher 8", "Mr", new int[]{1,1,0,0,1,1,1});
t[8] = new Teacher("Teacher 9", "Mr", new int[]{1,1,1,1,1,0,0});
t[9] = new Teacher("Teacher 10", "Mr", new int[]{0,0,0,1,1,1,0});
t[10] = new Teacher("Teacher 11", "Mr", new int[]{0,0,1,0,0,1,1});
t[11] = new Teacher("Teacher 12", "Mr", new int[]{0,0,1,1,0,1,0});
t[12] = new Teacher("Teacher 13", "Mr", new int[]{1,1,1,1,0,0,0});
t[13] = new Teacher("Teacher 14", "Mr", new int[]{1,1,0,1,1,0,1});
以上的输出(使用我正在使用的算法):
P1 P2 P3 P4 P5 P6 P7
Teacher 1 1 - 1 - - - -
Teacher 2 2 1 - 1 1 - 1
Teacher 3 - 2 2 2 2 2 -
Teacher 4 3 3 - 3 3 - 3
Teacher 5 4 4 - - 4 3 4
Teacher 6 5 5 4 - - 4 5
Teacher 7 - - 5 - 5 5 6
Teacher 8 6 6 - - 6 6 7
Teacher 9 7 7 6 7 7 - -
Teacher 10 - - - 8 8 7 -
Teacher 11 - - 8 - - 8 8
Teacher 12 - - 9 9 - 9 -
Teacher 13 8 9 10 10 - - -
Teacher 14 9 10 - 11 9 - 10
如您所见,程序只是循环遍历有效空间,用 subs 填充它们,直到 sub 达到其最大教学周期,然后开始一个新的 sub。问题是,我已经能够在手动操作时将使用的潜艇数量减少到 10 个,所以我一直在尝试寻找更有效的算法,但无济于事。
对于此输入,使用的最小潜艇数量为 9(受 P2 列约束),所以我想看看是否有任何可能的方法可以完成该数字,或者至少 10 个潜艇。提前致谢!