概述
我有一群人,他们每个人都有朋友名单和敌人名单.我想把它们排成一行(桌子上没有圆圈),所以没有敌人,只有朋友彼此相邻.
输入示例：https://gist.github.com/solars/53a132e34688cc5f396c
我想我需要使用图形着色来解决这个问题,但我不确定如何 – 我想我必须省略朋友(或敌人)列表以使其更容易并映射到图表.
有谁知道如何解决这些问题,并告诉我,如果我走在正确的道路上？
代码示例或在线示例也不错,我不介意编程语言,我通常使用Ruby,Java,Python,Javascript
非常感谢你的帮助！
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.
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