Cheapest Flights Within K Stops — C++

// Time:  O((|E| + |V|) * log|V|) = O(|E| * log|V|)
// Space: O(|E| + |V|) = O(|E|)

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
        using P = pair<int, int>;
        unordered_map<int, vector<P>> adj;
        for (const auto& flight : flights) {
            int u, v, w;
            tie(u, v, w) = make_tuple(flight[0], flight[1], flight[2]);
            adj[u].emplace_back(v, w);
        }
        
        unordered_map<int, unordered_map<int, int>> best;
        best[src][K + 1] = 0;
        using T = tuple<int, int, int>;
        priority_queue<T, vector<T>, greater<T>> min_heap;
        min_heap.emplace(0, src, K + 1);
        while (!min_heap.empty()) {
            int result, u, k;
            tie(result, u, k) = min_heap.top(); min_heap.pop();
            if (k < 0 ||
                (best.count(u) && best[u].count(k) &&  best[u][k] < result)) {
                continue;
            }
            if (u == dst) {
                return result;
            }
            for (const auto& kvp : adj[u]) {
                int v, w;
                tie(v, w) = kvp;
                if (!best.count(v) ||
                    !best[v].count(k - 1) ||
                    result + w < best[v][k - 1]) {
                    best[v][k - 1] = result + w;
                    min_heap.emplace(result + w, v, k - 1);
                }
            }
        }
        return -1;
    }
};