#1) Brantley College
You are looking for a minimal span solution because you have to visit all nodes
(buildings) in the network. Draw a map of the buildings and the distances between them.
It should look something like this:
Beginning at node 1, proceed through the network, choosing the shortest segment leaving
each node, provided that it does not lead to a node that has already been visited.
The shortest path to an unvisited node leaving Node 1 goes to Node 2, with a length of 3.
The shortest path to an unvisited node leaving Node 2 goes to Node 4, with a length of 2.
The shortest path to an unvisited node leaving Node 4 goes to Node 6, with a length of 3.
Up to this point the shortest path is 1
2
4
6, with a total length of 800 feet.
There are two paths leaving Node 6 with equal lengths, one going to Node 3 and one
going to Node 5. Each of these paths has a length of 4.
The two paths are now:
1
2
4
6
3, with a total length of 1200 feet
1
2
4
6
5, with a total length of 1200 feet
For the first of these two paths, the only remaining unvisited node is Node 5, with a
length of 4, making the final path
1
2
4
6
3
5, with a total length of 1600 feet.
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For the second of these two paths, the only remaining unvisited node is Node 3, with a
length of 4, making the final path
1
2
4
6
5
3, with a total length of 1600 feet.
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 Spring '11
 Das
 Accounting, Flow network, Maxflow mincut theorem, Mark Node

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