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.