Difference between revisions of "TADM2E 6.21"

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(Replaced content with "China get out")
(Undo revision 774 by FuckYou (talk))
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China get out
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Step 1:
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Perform topological sorting of the graph (we can do it as Graph is acyclic). This is O(n + m)
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Step 2:
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Go through vertices in topological order. Initially all vertices marked as inaccessible and only starting vertex marked with 0 distance.
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For every vertex in cycle, if it is accessible, we update all its children with new distance if new distance is shorter then previous one.
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Topological sorting ensures that we don't ever need to go backtrack. This is O(n + m).
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Total running bound is O(n+m), which is linear as requested.

Revision as of 10:44, 23 July 2020

Step 1:

Perform topological sorting of the graph (we can do it as Graph is acyclic). This is O(n + m)

Step 2:

Go through vertices in topological order. Initially all vertices marked as inaccessible and only starting vertex marked with 0 distance. For every vertex in cycle, if it is accessible, we update all its children with new distance if new distance is shorter then previous one. Topological sorting ensures that we don't ever need to go backtrack. This is O(n + m).

Total running bound is O(n+m), which is linear as requested.