Single source shortest distances in O(V+E) time for DAG (Directed Acyclic Graph). The idea is to use Topological Sorting.
Algorithm :
1. Initialize dist[] = {INF, INF, ….} and dist[s] = 0 where s is
the source vertex.
2. Create a toplogical order of all vertices.
3. Do following for every vertex u in topological order.
Do following for every adjacent vertex v of u
if (dist[v] > dist[u] + weight(u, v))
dist[v] = dist[u] + weight(u, v)
Time Complexity: Time complexity of topological sorting is O(V+E). After finding topological order, the algorithm process all vertices and for every vertex, it runs a loop for all adjacent vertices. Total adjacent vertices in a graph is O(E). So the inner loop runs O(V+E) times. Therefore, overall time complexity of this algorithm is O(V+E).
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