How do you compute the total weight of the MST after running Prim's algorithm?

When multiple candidate edges share the same minimum weight, you can pick any one of them. The choice does not affect the validity of the greedy step, though it may lead to a different MST if edge weights are not distinct.

Conditions: Multiple edges connect the visited set to unvisited nodes.; These edges have identical minimum weights.

The time complexity of Prim's algorithm depends on the data structures used. Using an adjacency matrix with searching gives $O(V^2)$, while using a binary heap with an adjacency list gives $O(V \log V + E \log V)$, where $V$ is the number of vertices and $E$ is the number of edges.

Conditions: The graph is represented using specific data structures.; $V$ denotes the number of vertices.; $E$ denotes the number of edges.

Edge B-E is skipped because both of its endpoints (B and E) are already in the visited set (MST). Prim's algorithm only considers edges that connect a visited vertex to an unvisited vertex to avoid creating cycles.

Conditions: Both endpoints of the edge are already marked as visited.; The algorithm strictly follows the rule of adding only crossing edges.

No, the minimum spanning tree is not always unique. It is unique only if all edge weights in the graph are distinct.

Conditions: The graph is weighted.; Uniqueness is guaranteed only when all edge weights are pairwise distinct.

To start Prim's algorithm, create an empty list called 'visited' to track touched nodes, then pick an arbitrary starting node and add it to the visited list.

Conditions: The graph is weighted and connected.; You have a mechanism to track visited vertices (e.g., a list or set).