Original link: https://blog.csdn.net/WerKeyTom_FTD/java/article/details/65658944
hall theorem is about determining whether there is a perfect match for the bipartite graph.
Let's have some basic definitions.
The basic definition
is not well defined. .
Anyone who has studied network streaming should understand what is mentioned in this article.
Perfect matching means that the maximum matching number is min (| X |, | Y |),
which means that all points in one set of X or Y sets are matched.
Theorem content
Let's assume that the X set point is a little less. The X set is considered to have n points.
Then there is a perfect match in the bipartite graph G, then any positive integer 1 <= k <= n is satisfied, which satisfies that I select k different points from the X set, then the number of points in the y set to which they are connected is not less than k.
It is necessary to prove that
if a bipartite graph G has a perfect match and does not satisfy Hall's theorem.
Then for a certain k points, they are connected to less than k points.
So how are they all matched? ? ?
Obviously the necessity is correct.
Sufficiency proves that
if a bipartite graph G does not have a perfect match, and meet Hall's theorem.
So if there is a maximum matching scheme, since there is no perfect match, at least one unmatched point A can be found.
Because this bipartite graph satisfies Hall's theorem, this point must be connected to at least one point B (there may be multiple points).
If this point B is not in the maximum match, they match, how is it possible? ? ?
Then this point B is in the maximum match! So there must be a point C on the left that matches it.
As a matching point, C may find D on the right, so keep searching, because the left point is <= the right point,
so the final point falls on the right point and ends, finding a Zengguang Road.
So there was a contradiction.
C may also find another matching point on the right, because C only matches B, and the shape is as follows:
At this time, we can delete the two points B and C, which does not affect the solution of the problem, let A go to find D, and then just like the above proof.