No. Discovery is tentative. In Dijkstra, extracting the current minimum valid entry is the safe commitment under nonnegative edge weights. An old heap entry for cost 10 can be ignored after the improvement.

If the cheapest unsettled vertex has tentative distance d, a route through another unsettled vertex cannot later become cheaper by adding only nonnegative edges. More formally, a supposedly cheaper route has a first unsettled vertex whose settled predecessor would already have offered a tentative cost no larger than that route. That contradicts selecting a larger minimum.

Use integer vertex labels and adjacency lists of (neighbor, weight), assuming all weights are nonnegative.

from heapq import heappop, heappush

def cheapest_distances(graph, start):
    distances = {start: 0}
    heap = [(0, start)]
    while heap:
        cost, node = heappop(heap)
        if cost != distances[node]:
            continue
        for neighbor, weight in graph.get(node, []):
            candidate = cost + weight
            if candidate < distances.get(neighbor, float("inf")):
                distances[neighbor] = candidate
                heappush(heap, (candidate, neighbor))
    return distances

graph = {0: [(1, 10), (2, 1)], 2: [(1, 2)], 1: []}
assert cheapest_distances(graph, 0) == {0: 0, 1: 3, 2: 1}
assert cheapest_distances({}, 0) == {0: 0}

Relaxation means offering a better route through the current node. This implementation leaves outdated offers in the heap and skips them when popped. Its general bound is O((V+E) log(V+E)) time and O(V+E) extra space; for simple graphs the customary time bound is O((V+E) log V).

Analogy limit: this method assumes nonnegative edges. Adding a visited set or a heap does not make negative-edge shortest paths safe under the same proof. DAG shortest-path evaluation or Bellman–Ford may fit different conditions.

No. BFS’s queue already processes the correct distance order in O(V+E) time. More general machinery is unnecessary for that uniform-cost contract.