Here the graph maps each node to a list of (neighbour, weight) pairs:

import heapq

def dijkstra(graph, start):
    dist = {start: 0}
    queue = [(0, start)]
    while queue:
        d, node = heapq.heappop(queue)
        if d > dist[node]:
            continue
        for nbr, weight in graph[node]:
            new = d + weight
            if new < dist.get(nbr, float("inf")):
                dist[nbr] = new
                heapq.heappush(queue, (new, nbr))
    return dist

roads = {
    "A": [("B", 4), ("C", 1)],
    "B": [("D", 1)],
    "C": [("B", 2), ("D", 5)],
    "D": [],
}
print(dijkstra(roads, "A"))

The direct road A to B is 4, but A to C to B is only 3, and Dijkstra finds it. The if d > dist[node]: continue line skips stale queue entries for nodes already reached more cheaply. float("inf") is infinity, larger than any number. (Dijkstra's algorithm needs weights that aren't negative.)