Dijkstra shortest path algorithm based on python heapq heap implementation
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from collections import defaultdict | |
from heapq import * | |
def dijkstra(edges, f, t): | |
g = defaultdict(list) | |
for l,r,c in edges: | |
g[l].append((c,r)) | |
q, seen, mins = [(0,f,())], set(), {f: 0} | |
while q: | |
(cost,v1,path) = heappop(q) | |
if v1 not in seen: | |
seen.add(v1) | |
path = (v1, path) | |
if v1 == t: return (cost, path) | |
for c, v2 in g.get(v1, ()): | |
if v2 in seen: continue | |
prev = mins.get(v2, None) | |
next = cost + c | |
if prev is None or next < prev: | |
mins[v2] = next | |
heappush(q, (next, v2, path)) | |
return float("inf"), None | |
if __name__ == "__main__": | |
edges = [ | |
("A", "B", 7), | |
("A", "D", 5), | |
("B", "C", 8), | |
("B", "D", 9), | |
("B", "E", 7), | |
("C", "E", 5), | |
("D", "E", 15), | |
("D", "F", 6), | |
("E", "F", 8), | |
("E", "G", 9), | |
("F", "G", 11) | |
] | |
print "=== Dijkstra ===" | |
print edges | |
print "A -> E:" | |
print dijkstra(edges, "A", "E") | |
print "F -> G:" | |
print dijkstra(edges, "F", "G") |
Nice and clean
Thank you very much for this beautiful algorithm.
pretty sure this is not Dijkstra; you're doing heappush(q, (next, v2, path))
at the very end, but in True dijkstra it would need a call to "decrease_key", which in python is heap._siftdown
@xdavidliu I was confused by this until I saw https://stackoverflow.com/a/31123108. I think Dijkstra's algorithm is a higher level concept, so either implementation is valid.
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Thank you so much for this gift, very clean and clever solution 😄
If anyone just wonders how to easily receive as output only the value of the solution remove the cost from the return at line 15:
if v1 == t: return cost
instead of
if v1 == t: return (cost, path)