787. Cheapest Flights Within K Stops
There are n cities connected by m flights. Each flight starts from city u and arrives at v with a price w.
Now given all the cities and flights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.
**Example 1:**
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
Output: 200
Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture.
**Example 2:**
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0
Output: 500
Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.Constraints:
- The number of nodes nwill be in range[1, 100], with nodes labeled from0ton- 1.
- The size of flightswill be in range[0, n * (n - 1) / 2].
- The format of each flight will be (src,dst, price).
- The price of each flight will be in the range [1, 10000].
- kis in the range of- [0, n - 1].
- There will not be any duplicated flights or self cycles.
# @lc code=start
using LeetCode
function find_cheapest_price(n::Int, flights::Vector{Vector{Int}}, src::Int, dst::Int, K::Int)
    graph = [Dict{Int, Int}() for i in 1:n]
    for flight in flights
        graph[flight[1]][flight[2]] = flight[3]
    end
    cost = fill(typemax(Int) >> 1, n)
    cost[src] = 0
    changed = [src]
    for i in 1:K+1
        tmp = Int[]
        for new_s in changed
            for (s_to_t, st_cost) in graph[new_s]
                if cost[s_to_t] > st_cost + cost[new_s]
                    cost[s_to_t] = st_cost + cost[new_s]
                    push!(tmp, s_to_t)
                end
            end
        end
        changed = tmp
    end
    return (cost[dst] == typemax(Int) >> 1) ? -1 : cost[dst]
end
# @lc code=endfind_cheapest_price (generic function with 1 method)
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