问题
Given a random unidirected graph, I must find 'bottleneck edges' to get from one vertex to another.
What I call 'bottleneck edges' (there must be a better name for that!) -- suppose I have the following unidirected graph:
A
/ | \
B--C--D
| |
E--F--G
\ | /
H
To get from A to H independently of the chosen path edges BE and DG must always be traversed, therefore making a 'bottleneck'.
Is there a polynomial time algorithm for this?
edit: yes, the name is 'minimum cut' for what I meant witch 'bottleneck edges'.
回答1:
Sounds like you need minimum cut, the minimal set of edges removing which will separate your graph into two pieces.
http://en.wikipedia.org/wiki/Minimum_cut
回答2:
What you are looking for is a cut. Given a graph, a cut is a set of edges that partitions the vertices into two disjoint subsets.
Assuming you are trying to get the smallest cut possible, this is the classic min-cut problem. Here is a pseudo code version of the Ford-fulkerson algorithm, reworked for your case (undirected, unweighted graphs). I'm pretty sure it should work, but I am not sure I'm being the most efficient / idiomatic here.
reorganize your graph into a directed graph,
with two directed edges (u->v, v->u) for each original edge (u-v)
while there is a path P from A to H:
(hint: use breadth first search to find paths - long story here)
//augment the path P:
for each edge (u->v) in P:
remove (u->v) from the graph and add (v->u) to it
(if v->u isn't there already)
Label all vertices as reacheable or not reacheable from A.
The bottleneck edges is the set of edges
that connect a reacheable and a unreacheable vertex
For example, in your case BFS would give us the path A-B-E-H. After removing these edges, we would still be able to find the path A-D-G-H. After these edges are removed, the graph is partitioned into the reacheable vertices {A,B,C,D} and the unreacheable {E,F,G,H}. The edges that have a vertex from each (B-E and D-G) set are the bottleneck edges.
来源：https://stackoverflow.com/questions/5810703/finding-bottleneck-edges-in-a-graph