有三种词典: 卡车、物品和装载的卡车
如何替换Trucks和Items字典中的truck键和值?你知道吗
truck_dic = {6: [21.0, 7.0, 7.0],
7: [23.0, 7.0, 7.0],
8: [27.0, 7.0, 7.0],
9: [20.0, 7.0, 7.0],
10: [24.0, 7.0, 7.0],
11: [28.0, 8.0, 8.0],
12: [32.0, 8.0, 8.0],
13: [32.0, 9.0, 10.0],
14: [32.0, 9.0, 10.0],
15: [20.0, 8.0, 8.0],
16: [20.0, 8.0, 8.0]}
Items_dic = {0: [4.6, 4.3, 4.3],
1: [4.6, 4.3, 4.3],
2: [6.0, 5.6, 9.0],
3: [8.75, 5.6, 6.6],
4: [6.0, 5.16, 6.6],
5: [6.0, 5.6, 9.0],
6: [8.75, 5.6, 6.6],
7: [4.6, 4.3, 4.3],
8: [6.0, 5.6, 9.0],
9: [8.75, 5.6, 6.6],
10: [6.0, 5.16, 6.6]}
loaded_trucks = {'23.0x7.0x7.0': ['4.60x4.30x4.30, 4.60x4.30x4.30,
6.60x6.00x5.16, 6.60x6.00x5.16'],
'27.0x7.0x7.0': ['4.60x4.30x4.30, 9.00x6.00x5.60,
8.75x6.60x5.60'],
'20.0x8.0x8.0': ['9.00x6.00x5.60, 8.75x6.60x5.60'],
'21.0x7.0x7.0': ['9.00x6.00x5.60, 8.75x6.60x5.60']}
以至于它
loaded_trucks = {'7': ['0', '7', '4', '10'],
'8': ['1', '5', '9'],
'16': ['2', '3'],
'6': ['6', '8']}
PS:装载的卡车尺寸混乱
输出
result
是请注意,具有相同尺寸的项目可以互换。例如,第5项和第8项的尺寸相同,因此可以互换。你知道吗
另外,我们也不关心是否重新排列了各个维度的顺序,因为这只意味着项目是旋转的。例如,尺寸为“4.60x4.30x4.30”的项目被视为与另一个尺寸为“4.30x4.30x4.60”的项目相同。你知道吗
最后,每个项只分配一次,但我们假设原始
loaded_trucks
字典中的项都在Items_dic
中,因此不会出现短缺或剩余项,否则从列表中弹出将产生错误。你知道吗相关问题 更多 >
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