There are n cities connected by some number of flights. You are given an array flights where flights[i] = [fromi, toi, pricei] indicates that there is a flight from city fromi to city toi with cost pricei.
You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.
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Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-3drawio.png\")
\n\nInput: n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1\nOutput: 700\nExplanation:\nThe graph is shown above.\nThe optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.\nNote that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.\n\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-1drawio.png\")
\n\nInput: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1\nOutput: 200\nExplanation:\nThe graph is shown above.\nThe optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.\n\n\n
Example 3:
\n![\"\"](\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-2drawio.png\")
\n\nInput: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0\nOutput: 500\nExplanation:\nThe graph is shown above.\nThe optimal path with no stops from city 0 to 2 is marked in red and has cost 500.\n\n\n
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Constraints:
\n\n1 <= n <= 1000 <= flights.length <= (n * (n - 1) / 2)flights[i].length == 30 <= fromi, toi < nfromi != toi1 <= pricei <= 1040 <= src, dst, k < nsrc != dstCompiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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