问题
Some pseudocode here (disregard my style)
Starting from v1(enqueued):
function BFS(queue Q)
v2 = dequeue Q
enqueue all unvisited connected nodes of v2 into Q
BFS(Q)
end // maybe minor problems here
Since there are V vertices in the graph, and these V vertices are connected to E edges, and visiting getting connected nodes (equivalent to visiting connected edges) is in the inner loop (the outer loop is the recursion itself), it seems to me that the complexity should be O(V*E) rather than O(V+E). Can anyone explain this for me?
回答1:
E is not the number of edges adjacent to each vertex - its actually the total number of edges in the graph. Defining it this way is useful because you don't necessarily have the same number of edges on every single vertex.
Since each edge gets visited once by the time the DFS ends, you get O(E) complexity from that part. Then you add the O(V) for visiting each vertex once and get O(V + E) on total.
来源：https://stackoverflow.com/questions/18604803/why-is-the-complexity-of-bfs-ove-instead-of-ove