Find Edges in Shortest Paths — C++

// Time:  O((|E| + |V|) * log|V|) = O(|E| * log|V|) by using binary heap,
//        if we can further to use Fibonacci heap, it would be O(|E| + |V| * log|V|)
// Space: O(|E| + |V|) = O(|E|)

// dijkstra's algorithm
class Solution {
public:
    vector<bool> findAnswer(int n, vector<vector<int>>& edges) {
        static const int INF = numeric_limits<int>::max();

        vector<vector<pair<int, int>>> adj(n);
        for (const auto& e : edges) {
            adj[e[0]].emplace_back(e[1], e[2]);
            adj[e[1]].emplace_back(e[0], e[2]);
        }
        const auto& dijkstra = [&](int start) {
            vector<int64_t> best(size(adj), INF);
            best[start] = 0;
            priority_queue<pair<int64_t, int>, vector<pair<int64_t, int>>, greater<pair<int64_t, int>>> min_heap;
            min_heap.emplace(0, start);
            while (!empty(min_heap)) {
                const auto [curr, u] = min_heap.top(); min_heap.pop();
                if (curr > best[u]) {
                    continue;
                }
                for (const auto& [v, w] : adj[u]) {
                    if (best[v] - curr <= w) {
                        continue;
                    }
                    best[v] = curr + w;
                    min_heap.emplace(best[v], v);
                }
            }
            return best;
        };

        const auto& dist1 = dijkstra(0);
        const auto& dist2 = dijkstra(n - 1);
        vector<bool> result(size(edges));
        int i = 0;
        for (const auto& e : edges) {
            result[i++] = (dist1[e[0]] != INF && dist2[e[1]] != INF && dist1[e[0]] + e[2] + dist2[e[1]] == dist1[n - 1]) ||
                          (dist2[e[0]] != INF && dist1[e[1]] != INF && dist2[e[0]] + e[2] + dist1[e[1]] == dist2[0]);
        } 
        return result;
    }
};