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我對下列問題的工作,現在的客戶端,這將創造最經濟的時間表(使用最少的替代品)說:替代調度程序/算法
- 替代品應該在的地方老師的工作作爲連續時間 越好(*不是一個巨大的關注)
- 替補可以爲6個週期
到目前爲止,我有一個老師級(如下圖所示),並實際創建一個管理類只工作最佳時間表。現在我只是讓程序循環遍歷網格來填充每個替代品。
Teacher[] t= new Teacher[14];
Organizer o = new Organizer(t);
o.sort();
int[][] g = o.getGrid();
例輸入:
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});
輸出,用於上述(與算法,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,程序對面的有效空間循環,將他們填入潛艇,直到潛艇達到最大教學時間,然後開始一個新的潛艇。問題是,當我手動完成時,我已經可以將使用的次數減少到10次,所以我一直在試圖找到一個更高效的算法,但沒有用。
對於此輸入,使用的最小子數爲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該數目,或至少10個子數。提前致謝!