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787. Cheapest Flights Within K Stops
Bookmark
BFS Levels
Bellman-Ford
Dijkstra + State
Input
4 Nodes (k=1)
3 Nodes (k=1)
5 Nodes (k=2)
Custom
n
=
4
,
flights
=
[[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]]
,
src
=
0
,
dst
=
3
,
k
=
1
Flight Network
100
100
100
600
200
0
1
2
3
Frontier
empty
dist
city 0
INF
city 1
INF
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
Frontier
empty
dist
city 0
INF
city 1
INF
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
Frontier starts at city 0
Frontier
0
dist
city 0
0
city 1
INF
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
Level 1 of 2, frontier [0]
Frontier
0
dist
city 0
0
city 1
INF
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
0 + 100 = 100 vs INF
Frontier
0
dist
city 0
0
city 1
100
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
Next frontier [1]
Frontier
1
dist
city 0
0
city 1
100
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
Level 2 of 2, frontier [1]
Frontier
1
dist
city 0
0
city 1
100
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
100 + 100 = 200 vs INF
Frontier
1
dist
city 0
0
city 1
100
city 2
INF
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
100 + 600 = 700 vs INF
Frontier
2
dist
city 0
0
city 1
100
city 2
200
city 3
INF
Flight Network
100
100
100
600
200
0
1
2
3
100 + 600 = 700 vs INF
Frontier
2
3
dist
city 0
0
city 1
100
city 2
200
city 3
700
Flight Network
100
100
100
600
200
0
1
2
3
Cheapest route to 3 costs 700
Frontier
empty
dist
city 0
0
city 1
100
city 2
200
city 3
700
algo
master
.
io
Step:
4 cities and 5 flights. Find the cheapest route from 0 to 3 using at most 1 stop(s).
0 / 20
Input
4 Nodes (k=1)
3 Nodes (k=1)
5 Nodes (k=2)
Custom
n
=
4
,
flights
=
[[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]]
,
src
=
0
,
dst
=
3
,
k
=
1
0 / 20
algo
master
.
io
Step:
4 cities and 5 flights. Find the cheapest route from 0 to 3 using at most 1 stop(s).
