为什么DFS和BFS的时间复杂度取决于图表的表示方式？

Question

The site http://web.eecs.utk.edu/~huangj/CS302S04/notes/graph-searching.html describes that when an adjacency list is used then, DFS and BFS have complexity O(V+E), and if an adjacency matrix is used, the complexity is O(V ² ). Why is this?

Answer 1

In both cases, the runtime depends on how long it takes to iterate across the outgoing edges of a given node. With an adjacency list, the runtime is directly proportional to the number of outgoing edges. Since each node is visited once, the cost is the number of nodes plus the number of edges, which is O(m + n). With am adjacency matrix, the time required to find all outgoing edges is O(n) because all n columns in the row for a node must be inspected. Summing up across all n nodes, this works out to O(n ² ).

Hope this helps!

Answer 2

你必须注意，为了探索探索所需的每个顶点时间，它只等于c * x，其中x是顶点的indegree。由于我们有兴趣找到整体复杂性，总时间将是c1 * x1 + c2 * 对于n个节点， x2 ... cn xn。看到max（ci）= d，我们看到总时间<= d （所有顶点的不一致之和）= d * 2m = O（m）。这里我们已经计算过不是一个顶点的时间，而是所有顶点合在一起的时间。但是排队操作需要时间O（n），所以总体上是O（n + m）。

Answer 3

The time complexity for both DFS and BFS can be computed as follows:

Iterating every vertex once and its corresponding incident edges , so the total time complexity will be ->

Time Complexity = v1 + (incident_edges on v1) + v2 + (incident_edges on v2) + ...... + vn + ( incident_edges on vn)

Now this can be regrouped as -> (v1+v2+v3+.....vn) + (incident_edges on v1 + incident_edges on v2 + ..... incident_edges on vn)

Thus total time complexity would turn out to be = (v1+v2+v3+.....vn) + (incident_edges on v1 + incident_edges on v2 + ..... incident_edges on vn)

(v1 + v2 + ... + vn) = V or n (Total number of vertices)

For adjacency list representation :

(incident_edges on v1 + incident_edges on v2 + ..... incident_edges on vn) = E(Total number of edges)

Thus for adjacency list representation time complexity will be O(V+E)

For adjacency matrix representation :

To visit the neighbors of the corresponding node(Row) we need to iterate all the columns for the particular row which amounts to V

So, (incident_edges on v1 + incident_edges on v2 + ..... incident_edges on vn) = V + V + .... Vth time V) = V*V

Thus time complexity will be O(V + V^2) = O(V^2)