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3112. Minimum Time to Visit Disappearing Nodes 👍

  • Time: $O((|V| + |E|)\log |V|)$
  • Space: $O(|E| + |V|)$
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class Solution {
 public:
  vector<int> minimumTime(int n, vector<vector<int>>& edges,
                          vector<int>& disappear) {
    vector<vector<pair<int, int>>> graph(n);

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

    return dijkstra(graph, 0, disappear);
  }

 private:
  vector<int> dijkstra(const vector<vector<pair<int, int>>>& graph, int src,
                       const vector<int>& disappear) {
    vector<int> dist(graph.size(), INT_MAX);

    dist[src] = 0;
    using P = pair<int, 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;
      for (const auto& [v, w] : graph[u])
        if (d + w < disappear[v] && d + w < dist[v]) {
          dist[v] = d + w;
          minHeap.push({dist[v], v});
        }
    }

    for (int& d : dist)
      if (d == INT_MAX)
        d = -1;

    return dist;
  }
};

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class Solution {
  public int[] minimumTime(int n, int[][] edges, int[] disappear) {
    List<Pair<Integer, Integer>>[] graph = new List[n];

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

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

    return dijkstra(graph, 0, disappear);
  }

  private int[] dijkstra(List<Pair<Integer, Integer>>[] graph, int src, int[] disappear) {
    int[] dist = new int[graph.length];
    Arrays.fill(dist, Integer.MAX_VALUE);

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

    while (!minHeap.isEmpty()) {
      final int d = minHeap.peek().getKey();
      final int u = minHeap.poll().getValue();
      if (d > dist[u])
        continue;
      for (Pair<Integer, Integer> pair : graph[u]) {
        final int v = pair.getKey();
        final int w = pair.getValue();
        if (d + w < disappear[v] && d + w < dist[v]) {
          dist[v] = d + w;
          minHeap.offer(new Pair<>(dist[v], v));
        }
      }
    }

    for (int i = 0; i < dist.length; ++i)
      if (dist[i] == Integer.MAX_VALUE)
        dist[i] = -1;

    return dist;
  }
}

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class Solution:
  def minimumTime(
      self,
      n: int,
      edges: list[list[int]],
      disappear: list[int],
  ) -> list[int]:
    graph = [[] for _ in range(n)]

    for u, v, w in edges:
      graph[u].append((v, w))
      graph[v].append((u, w))

    return self._dijkstra(graph, 0, disappear)

  def _dijkstra(
      self,
      graph: list[list[tuple[int, int]]],
      src: int,
      disappear: list[int],
  ) -> list[int]:
    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
      for v, w in graph[u]:
        if d + w < disappear[v] and d + w < dist[v]:
          dist[v] = d + w
          heapq.heappush(minHeap, (dist[v], v))

    return [d if d != math.inf else -1
            for d in dist]