Question

我必须使用基于最小堆的优先级队列来实现Prim的算法。如果我的图包含带有以下undirected邻接列表的顶点A，B，C和D ...... [它被排序为（顶点名称，相邻顶点的权重）]

A -> B,4 -> D,3
B -> A,4 -> C,1 -> D,7
C -> B,1
D -> B,7 -> A,3

粗略图：

A-4-B-1-C
|   /
3  7
| /
D

优先级队列是什么样的？我不知道应该把它放进去。我应该放一切吗？我应该只使用A B C和D.我没有任何线索，我真的想要一个答案。

Answer 1

Prim's: grow the tree by adding the edge of min weight with exactly one end in the tree.
The PQ contains the edges with one end in the tree.
Start with vertex 0 added to tree and add all vertices connected to 0 into the PQ.
DeleteMin() will give you the min weight edge (v, w), you add it to the MST and add all vertices connected to w into the PQ.

is this enough to get you started?
---
so, in your example, the in the first iteration, the MST will contain vertex A, and the PQ will contain the 2 edges going out from A:
A-4-B
A-3-D

Answer 2

这是prim的算法：

Choose a node. 
Mark it as visited.
Place all edges from this node into a priority queue (sorted to give smallest weights first).  
While queue not empty:  
    pop edge from queue
    if both ends are visited, continue
    add this edge to your minimum spanning tree
    add all edges coming out of the node that hasn't been visited to the queue
    mark that node as visited

所以为了回答你的问题，你将边缘放在一个节点上。

如果将所有边缘放入优先级队列，那么你就拥有了Kruskal的算法，该算法也用于最小生成树。

这取决于您如何表示图表的运行时间。邻接列表使得Kruskal和Prim的复杂度O（E log E）为O（E log V），除非您使用斐波那契堆，在这种情况下您可以实现O（E + V log V）。

Answer 3

您可以为顶点指定权重。然后根据这些权重使用优先级队列。这是来自wiki的参考：http://en.wikipedia.org/wiki/Prim's_algorithm

MST-PRIM (G, w, r) {
for each u ∈ G.V
u.key = ∞
u.parent = NIL
r.key = 0
Q = G.V
while (Q ≠ ø)
u = Extract-Min(Q)
for each v ∈ G.Adj[u]
if (v ∈ Q) and w(u,v) < v.key
v.parent = u
v.key = w(u,v)
}

Q将是您的优先级队列。您可以使用struct来保存顶点的信息。

prim的算法解释

3 个答案: