相关疑难解决方法(0)

BFS的基本算法:

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

所以我认为时间的复杂性将是:

v1 + (incident edges) + v2 + (incident edges) + .... + vn + (incident edges) 

v顶点1到哪里n

首先,我说的是正确的吗？其次,这是怎样的O(N + E),直觉为什么会非常好.谢谢