Find weight of MST in a complete graph of two weight with given edges

Question

I need to get the MST of a complete graph where all edges are defaulted to weight 3, and I'm also given edges that have weight 1.

Here is an example

5 4 (N, M)
1 5
1 4
4 2
4 3

Resulting MST = 3 -> 5 -> 1 -> 4 -> 2

Where the first row has the number of total nodes (N), the amount of 1-weight edges (M) and all of the following (M) rows contain an edge that has weight of 1.

I tried constructing a complete graph and updating weights of given edges to 1, but the spacial complexity is too big for a problem that contains up to 10^5 1-weight edges.

score 2 · Accepted Answer · answered Nov 28 '21 at 22:49

2

Use Kruskal's algorithm to construct the minimum spanning forest of the graph of only the 1-weight edges. The total weight of the minimum spanning tree of the entire graph will then be the number of edges in the minimum spanning forest (times 1) plus the cost of the edges to connect the trees in the forest (number of trees minus 1 times the weight of 3) plus the cost of one 3-weight edge each to connect the remaining nodes not in the original minimum spanning forest.

answered Nov 28 '21 at 22:49

beaker

16,331
3
32
49

2

Reduces to 3(N-1) - 2(number of non-redundant 1-edges) – Matt Timmermans Nov 28 '21 at 22:55
Is this possible to do even if the 1 weight edges don't connect? For example if it has the edges 1 - 2 and 4 - 5. – JameEnder Nov 28 '21 at 22:56
1

@MattTimmermans nice :) – beaker Nov 28 '21 at 22:58
1

@JameEnder That's why I said 'forest'. If you have only 1-2 and 4-5, that's 2 trees in the forest that will need to be connected by a 3-weight edge. – beaker Nov 28 '21 at 22:59
@beaker So the MST of the said 'forest' is the MST of all weight 1 nodes connected + 3 * the number of weight 3 nodes needed to connect them? – JameEnder Nov 28 '21 at 23:03
1

@JameEnder that's right – beaker Nov 28 '21 at 23:04
@beaker Thank you! That's a solution that wouldn't cross my mind for the whole 3 days i've been working at it – JameEnder Nov 28 '21 at 23:05
@MattTimmermans Just to be clear, I'm assuming that by "non-redundant", you mean edges that do not create a cycle. – beaker Nov 28 '21 at 23:05
1

Yes -- the ones that Kruskal's would add to your forest. (I say Kruskal's, but of course you don't need to sort them) – Matt Timmermans Nov 28 '21 at 23:06

Find weight of MST in a complete graph of two weight with given edges

1 Answers1