Some knowledge of graph theory

The minimum vertex cover bipartite graph: In seeking the least bipartite graph edges, so that each edge wherein the at least one point and associated

Minimum Vertex Cover = largest number of matched

FIG DAG (directed graph without loops (Directed Acyclic Graph)) minimum path coverage: simple paths do not want to pay less exhausted cover all vertices in the graph

Vertices = minimum path cover - the maximum number of matches

No minimum coverage to the road map:

No path to cover most of the bipartite graph Vertices = - maximum bipartite matching / 2 (since there is no bi is equal to twice the side of a forward and reverse direction FIG FIGS, the resulting number of matches to be more than doubled so 2 is divided by the original number of matches)

The minimum point may be repeated walking path covered:

[Title] meaning: to send robots to Mars treasure hunt, gives a acyclic directed graph, the robot can land on any point, and then go to other points along the way to explore, our task is to calculate how much the robot can send at least access to all points. Some point to be repeated.

[Thinking]: We can still cover the problem into the smallest path. If a person needs to go through another point, when one came, so he flew directly over the point from the past, beyond the point directly down a point. If we give everyone this ability, the minimum path without repeating points covering the results obtained, it is the result of questions asked, because the place needs to be repeated as long as the fly past, you can not repeat it. This method gives the ability to the point of it all is to reach all points indirectly to direct reach.

Maximum independent set bipartite graph: In the bipartite graph is not adjacent to any two points of the largest set of vertices

= Maximum independent set of nodes - the maximum number of matches

FIG multiple matches of the two : a general bipartite graph matching only one point, in the multiple matching, a matching point may be more to the point

 

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Origin www.cnblogs.com/heimao5027/p/11234871.html