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我对下列问题的工作,现在的客户端,这将创造最经济的时间表(使用最少的替代品)说:替代调度程序/算法
- 替代品应该在的地方老师的工作作为连续时间 越好(*不是一个巨大的关注)
- 替补可以为6个周期
到目前为止,我有一个老师级(如下图所示),并实际创建一个管理类只工作最佳时间表。现在我只是让程序循环遍历网格来填充每个替代品。
Teacher[] t= new Teacher[14];
Organizer o = new Organizer(t);
o.sort();
int[][] g = o.getGrid();
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});
输出,用于上述(与算法,我使用):
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
正如你所看到的,程序对面的有效空间循环,将他们填入潜艇,直到潜艇达到最大教学时间,然后开始一个新的潜艇。问题是,当我手动完成时,我已经可以将使用的次数减少到10次,所以我一直在试图找到一个更高效的算法,但没有用。
对于此输入,使用的最小子数为9(受P2列限制),所以我想看看是否有任何可能的方法可以完成该数目,或至少10个子数。提前致谢!