评论中已经提到,这个问题相当于旅行商问题.我想详细说明一下：

每个人都相当于一个顶点,并且边缘位于顶点之间,这些顶点代表可以相互对接的人.现在,找到可能的座位安排相当于在图中找到哈密顿路径.

所以这个问题就是NPC.最天真的解决方案是尝试所有可能的排列,从而导致O(n！)运行时间.有许多众所周知的方法比O(n！)表现更好,并且可以在网上免费获得.我想提一下Held-Karp,它运行在O(n ^ 2 * 2 ^ n)并且非常直接代码,这里是python：

#graph[i] contains all possible neighbors of the i-th person

def held_karp(graph):

n = len(graph)#number of persons

#remember the set of already seated persons (as bitmask) and the last person in the line

#thus a configuration consists of the set of seated persons and the last person in the line

#start with every possible person:

possible=set([(2**i,i) for i in xrange(n)])

#remember the predecessor configuration for every possible configuration:

preds=dict([((2**i,i),(0,-1)) for i in xrange(n)])

#there are maximal n persons in the line - every iterations adds a person

for _ in xrange(n-1):

next_possible=set()

#iterate through all possible configurations

for seated,last in possible:

for neighbor in graph[last]:

bit_mask=2**neighbor

if (bit_mask&seated)==0: #this possible neighbor is not yet seated!

next_config=(seated|bit_mask,neighbor)#add neighbor to the bit mask of seated

next_possible.add(next_config)

preds[next_config]=(seated,last)

possible=next_possible

#now reconstruct the line

if not possible:

return []#it is not possible for all to be seated

line=[]

config=possible.pop() #any configuration in possible has n person seated and is good enough!

while config[1]!=-1:

line.insert(0,config[1])

config=preds[config]#go a step back

return line

免责声明：此代码未经过适当测试,但我希望您能得到它的要点.