Medium
Minimum Time to Reach Destination in Directed Graph — C++
Full explanation · Time O(n + elogn) · Space O(n + e)
// Time: O(n + elogn)
// Space: O(n + e)
// dijkstra's algorithm
class Solution {
public:
int minTime(int n, vector<vector<int>>& edges) {
vector<vector<tuple<int, int, int>>> adj(n);
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1], e[2], e[3]);
}
const auto& dijkstra = [&]() {
vector<int> best(size(adj), numeric_limits<int>::max());
best[0] = 0;
using Data = pair<int, int>;
priority_queue<Data, vector<Data>, greater<Data>> min_heap;
min_heap.emplace(best[0], 0);
while (!empty(min_heap)) {
const auto [curr, u] = min_heap.top(); min_heap.pop();
if (curr != best[u]) {
continue;
}
if (u == size(adj) - 1) {
return curr;
}
for (const auto& [v, s, e] : adj[u]) {
if (curr > e) {
continue;
}
if (!(best[v] > max(curr, s) + 1)) {
continue;
}
best[v] = max(curr, s) + 1;
min_heap.emplace(best[v], v);
}
}
return -1;
};
return dijkstra();
}
};