Create minimum-cost-to-reach-city-with-discounts.cpp

Author: kamyu104

C++/minimum-cost-to-reach-city-with-discounts.cpp

@@ -0,0 +1,48 @@
+// 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|)
+
+class Solution {
+public:
+    int minimumCost(int n, vector<vector<int>>& highways, int discounts) {
+        using P = pair<int, int>;
+        unordered_map<int, vector<P>> adj;
+        for (const auto& highway : highways) {
+            int u, v, w;
+            tie(u, v, w) = make_tuple(highway[0], highway[1], highway[2]);
+            adj[u].emplace_back(v, w);
+            adj[v].emplace_back(u, w);
+        }
+        
+        unordered_map<int, unordered_map<int, int>> best;
+        best[0][discounts] = 0;
+        using T = tuple<int, int, int>;
+        priority_queue<T, vector<T>, greater<T>> min_heap;
+        min_heap.emplace(0, 0, discounts);
+        while (!empty(min_heap)) {
+            auto [total, u, k] = min_heap.top(); min_heap.pop();
+            if ((best.count(u) && best[u].count(k) &&  best[u][k] < total)) {
+                continue;
+            }
+            if (u == n - 1) {
+                return total;
+            }
+            for (const auto& [v, w] : adj[u]) {
+                if (!best.count(v) ||
+                    !best[v].count(k) ||
+                    total + w < best[v][k]) {
+                    best[v][k] = total + w;
+                    min_heap.emplace(total + w, v, k);
+                }
+                if (k > 0 &&
+                    (!best.count(v) ||
+                    !best[v].count(k - 1) ||
+                    total + w / 2 < best[v][k - 1])) {
+                    best[v][k - 1] = total + w / 2;
+                    min_heap.emplace(total + w / 2, v, k - 1);
+                }
+            }
+        }
+        return -1;
+    }
+};