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2473. Minimum Cost to Buy Apples 👍

  • Time: $O(|E| + |V|^2\log |E|)$
  • Space: $O(|V| + |E|)$
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class Solution {
 public:
  vector<long long> minCost(int n, vector<vector<int>>& roads,
                            vector<int>& appleCost, int k) {
    vector<long long> ans;
    vector<vector<pair<int, long>>> graph(n);

    for (const vector<int>& road : roads) {
      const int u = road[0] - 1;
      const int v = road[1] - 1;
      const int w = road[2];
      graph[u].emplace_back(v, w);
      graph[v].emplace_back(u, w);
    }

    for (int i = 0; i < n; ++i)
      ans.push_back(dijkstra(graph, i, appleCost, k));

    return ans;
  }

 private:
  long dijkstra(const vector<vector<pair<int, long>>>& graph, int src,
                const vector<int>& appleCost, int k) {
    long ans = LONG_MAX;
    vector<long> dist(graph.size(), LONG_MAX);

    dist[src] = 0;
    using P = pair<long, int>;  // (d, u)
    priority_queue<P, vector<P>, greater<>> minHeap;
    minHeap.emplace(dist[src], src);

    while (!minHeap.empty()) {
      const auto [d, u] = minHeap.top();
      minHeap.pop();
      if (d > dist[u])
        continue;
      ans = min(ans, appleCost[u] + (k + 1) * d);
      for (const auto& [v, w] : graph[u])
        if (d + w < dist[v]) {
          dist[v] = d + w;
          minHeap.emplace(dist[v], v);
        }
    }

    return ans;
  }
};

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class Solution {
  public long[] minCost(int n, int[][] roads, int[] appleCost, int k) {
    long[] ans = new long[n];
    List<Pair<Integer, Integer>>[] graph = new List[n];

    for (int i = 0; i < n; ++i)
      graph[i] = new ArrayList<>();

    for (int[] road : roads) {
      final int u = road[0] - 1;
      final int v = road[1] - 1;
      final int w = road[2];
      graph[u].add(new Pair<>(v, w));
      graph[v].add(new Pair<>(u, w));
    }

    for (int i = 0; i < n; ++i)
      ans[i] = dijkstra(graph, i, appleCost, k);

    return ans;
  }

  private long dijkstra(List<Pair<Integer, Integer>>[] graph, int i, int[] appleCost, int k) {
    long ans = Long.MAX_VALUE;
    long[] dist = new long[graph.length];
    Arrays.fill(dist, Long.MAX_VALUE);

    dist[i] = 0;
    Queue<Pair<Long, Integer>> minHeap = new PriorityQueue<>(Comparator.comparing(Pair::getKey)) {
      { offer(new Pair<>(dist[i], i)); } // (d, u)
    };

    while (!minHeap.isEmpty()) {
      final long d = minHeap.peek().getKey();
      final int u = minHeap.poll().getValue();
      if (d > dist[u])
        continue;
      ans = Math.min(ans, appleCost[u] + (k + 1) * d);
      for (Pair<Integer, Integer> pair : graph[u]) {
        final int v = pair.getKey();
        final int w = pair.getValue();
        if (d + w < dist[v]) {
          dist[v] = d + w;
          minHeap.offer(new Pair<>(dist[v], v));
        }
      }
    }

    return ans;
  }
}

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class Solution:
  def minCost(
      self,
      n: int,
      roads: list[list[int]],
      appleCost: list[int],
      k: int,
  ) -> list[int]:
    graph = [[] for _ in range(n)]

    for u, v, w in roads:
      graph[u - 1].append((v - 1, w))
      graph[v - 1].append((u - 1, w))

    return [self._dijkstra(graph, i, appleCost, k) for i in range(n)]

  def _dijkstra(
      self,
      graph: list[list[tuple[int, int]]],
      src: int,
      appleCost: list[int],
      k: int
  ) -> int:
    ans = math.inf
    dist = [math.inf] * len(graph)

    dist[src] = 0
    minHeap = [(dist[src], src)]  # (d, u)

    while minHeap:
      d, u = heapq.heappop(minHeap)
      if d > dist[u]:
        continue
      ans = min(ans, appleCost[u] + (k + 1) * d)
      for v, w in graph[u]:
        if d + w < dist[v]:
          dist[v] = d + w
          heapq.heappush(minHeap, (dist[v], v))

    return ans