我们可以找到一个瓶颈最小生成树为O(E日志* V)在最坏的情况下,通过使用Kruskal算法。这是因为每一个最小生成树是最起码的瓶颈生成树。
不过,我就死在从这个的过程。
我们如何才能找到一个最小的瓶颈,即使在最坏的情况下生成树的线性时间。请注意,我们可以假设,我们可以计算个密钥的中值线性时间在最坏的情况下
解决方案 获取 V
中,权重的中值| E |边缘。
找到所有边缘的值不大于 V
更多,并获得子图
如果该子连接, V
是答案的上限,并降低 V
,重复在步骤1,2。
如果子图没有连接,让连接的组件成为一个节点,并增加 V
,重复第1步,2
然后就可以解决线性时间的问题。
PS:使用DFS来判断子相连。与复杂度为O(E / 2)+ O(E / 4)+ O(E / 8)+ ... = O(E)
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We can find a minimum bottleneck spanning tree in O(E log*V) in the worst case by using Kruskal's algorithm. This is because every minimum spanning tree is a minimum bottleneck spanning tree.
But I got stuck on this job-interview question from this course.
How can we find a minimum bottleneck spanning tree in linear time even in the worst case. Note that we can assume that we can compute the median of n keys in linear time in the worst case.
解决方案 Get
V
, the median of the weights of the |E| edges.
Find all edge's value no more than V
, and get the subgraph
If the subgraph is connected, V
is the upper limit of the answer, and decrease the V
, repeat the step 1, 2.
If the subgraph is not connected, let the connected component to become a node, and increase the V
, repeat the step 1, 2.
Then you can solve the question in linear time.
PS: Using DFS to judge the subgraph is connected. and the complexity is O(E/2) + O(E/4) + O(E/8) + ... = O(E)
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