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Dijkstra algorithm modification with exactly one relaxation on a directed graph where the weights of outgoing edges of a node are the same

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xskxzr
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Consider the most common Dijkstrastandard version of Dijkstra's algorithm on directed graphs. ItAssume it is known that the input digraph $G = (V, E)$ has the following property -: for all $v \in V$ the weightsweight of all outcomingoutgoing edges $vu$ areis the same.

How can one modify the DijkstraDijkstra's algorithm so that exactly one relaxation is done for every vertex $v \in V$? (i. e. decreasing of current minimal path length $d_{v}$ is done exactly one timeonce for every $v$).

Consider the most common Dijkstra algorithm on directed graphs. It is known that input digraph $G = (V, E)$ has the following property - for all $v \in V$ the weights of all outcoming edges $vu$ are same.

How can one modify the Dijkstra algorithm so that exactly one relaxation is done for every vertex $v \in V$? (i. e. decreasing of current minimal path length $d_{v}$ is done exactly one time for every $v$).

Consider the standard version of Dijkstra's algorithm on directed graphs. Assume it is known that the input digraph $G = (V, E)$ has the following property: for all $v \in V$ the weight of all outgoing edges $vu$ is the same.

How can one modify Dijkstra's algorithm so that exactly one relaxation is done for every vertex $v \in V$? (i. e. decreasing of current minimal path length $d_{v}$ is done exactly once for every $v$).

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