Algorithm 如何找到在不同联合之间连接不同节点的最佳方法？

说明：

共有1000个联合，每个联合包含x个节点（x是1~100之间的随机值）。 现在我们可以创建从联合体a中的一个节点到联合体B中的另一个节点的连接

规则是：

1. one node only accepts just one connection. 
2. The connection must be across different unions.

同样，尽可能多地创建此类连接

最后，可能还有几个节点无法连接，因为其他联合中没有其他可用节点

例如：

Union 1: a b c
Union 2: d e
Union 3: f g h i j

如果我们选择以下连接：

U1.a <-> U2.d
U1.b <-> U2.e
U1.c <-> U3.f

这样就没有节点了

因此，问题是：

我们如何设计算法来尝试找到使无连接节点最少的最佳解决方案？

这看起来像是一个图：

G=(V,E), 
V = {all nodes}, 
E={(u,v) | u and v not in same union}

例如，它可以通过

来解决，这相当于，其中输入多集合中的每个元素都是并集的长度。此外，这个问题总是可以通过在O（n）时间内运行的简单贪婪实现来解决。例如，考虑以下输入：

Union 1: a a a
Union 2: b b b b b b b b b
Union 3: c c c c c c c c c c
Union 4: d d d d d d d d d d 
Union 5: e e

简单的贪婪算法创建两个输出列表。对于每个联合（从元素最多的联合开始），联合的元素将添加到较短的输出列表中。结果是两个列表，如下所示：

c c c c c c c c c c b b b b b b b b b
d d d d d d d d d d a a a e e

下一步是从较长列表的末尾提取一些项目，并将它们添加到较短列表的开头。在此示例中，移动了两个

b

s：

c c c c c c c c c c b b b b b b b 
b b d d d d d d d d d d a a a e e

那么，它会一直有效吗？是的，唯一的例外是当一个联合体包含超过项目总数一半的项目时。在这种情况下，不会从较长的列表中移动任何项目

下面是python中的一个示例实现：

inputList = [['a','a','a'],
            ['b','b','b','b','b','b','b','b','b'],
            ['c','c','c','c','c','c','c','c','c','c'],
            ['d','d','d','d','d','d','d','d','d','d'],
            ['e','e']]
topCount = 0
botCount = 0
topList = []
botList = []

# sort the input in descending order based on length
inputList = sorted(inputList, key=lambda x:len(x), reverse=True)

# greedy partitioning into two output lists
for itemList in inputList:
    if topCount <= botCount:
        topList += itemList
        topCount += len(itemList)
    else:
        botList += itemList
        botCount += len(itemList)

# move some elements from the end of the longer list to the beginning of the shorter list
if topList[0] != topList[-1]:
    if topCount > botCount+1:
        excess = (topCount - botCount) // 2
        botList = topList[-excess:] + botList
        topList = topList[:-excess]
    elif botCount > topCount+1:
        excess = (botCount - topCount) // 2
        topList = botList[-excess:] + topList
        botList = botList[:-excess]

print topList
print botList

---

我想你也应该试着问一下：

inputList = [['a','a','a'],
            ['b','b','b','b','b','b','b','b','b'],
            ['c','c','c','c','c','c','c','c','c','c'],
            ['d','d','d','d','d','d','d','d','d','d'],
            ['e','e']]
topCount = 0
botCount = 0
topList = []
botList = []

# sort the input in descending order based on length
inputList = sorted(inputList, key=lambda x:len(x), reverse=True)

# greedy partitioning into two output lists
for itemList in inputList:
    if topCount <= botCount:
        topList += itemList
        topCount += len(itemList)
    else:
        botList += itemList
        botCount += len(itemList)

# move some elements from the end of the longer list to the beginning of the shorter list
if topList[0] != topList[-1]:
    if topCount > botCount+1:
        excess = (topCount - botCount) // 2
        botList = topList[-excess:] + botList
        topList = topList[:-excess]
    elif botCount > topCount+1:
        excess = (botCount - topCount) // 2
        topList = botList[-excess:] + topList
        botList = botList[:-excess]

print topList
print botList

['c', 'c', 'c', 'c', 'c', 'c', 'c', 'c', 'c', 'c', 'b', 'b', 'b', 'b', 'b', 'b', 'b']
['b', 'b', 'd', 'd', 'd', 'd', 'd', 'd', 'd', 'd', 'd', 'd', 'a', 'a', 'a', 'e', 'e']