本文介绍了为什么DFS和时间复杂度BFS O(V + E)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
的基本算法BFS:
set start vertex to visited load it into queue while queue not empty for each edge incident to vertex if its not visited load into queue mark vertex
所以,我觉得时间复杂度将是:
So I would think the time complexity would be:
v1 + (incident edges) + v2 + (incident edges) + .... + vn + (incident edges)
其中, v 是顶点 1 到 N
首先,是我说的话是否正确?其次,这是怎么 O(N + E),和直觉至于为什么将是非常好的。谢谢
Firstly, is what I've said correct? Secondly, how is this O(N + E), and intuition as to why would be really nice. Thanks
推荐答案您总和
v1 + (incident edges) + v2 + (incident edges) + .... + vn + (incident edges)
可改写为
(v1 + v2 + ... + vn) + [(incident_edges v1) + (incident_edges v2) + ... + (incident_edges vn)]和第一组是 O(N)而另一个是O(E)。
and the first group is O(N) while the other is O(E).
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为什么DFS和时间复杂度BFS O(V + E)
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