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1976. Number of Ways to Arrive at Destination 👍

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

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

    return dijkstra(graph, 0, n - 1);
  }

 private:
  // Similar to 1786. Number of Restricted Paths From First to Last Node
  int dijkstra(const vector<vector<pair<int, int>>>& graph, int src, int dst) {
    constexpr int kMod = 1'000'000'007;
    vector<long> ways(graph.size());
    vector<long> dist(graph.size(), LONG_MAX);

    ways[src] = 1;
    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;
      for (const auto& [v, w] : graph[u])
        if (d + w < dist[v]) {
          dist[v] = d + w;
          ways[v] = ways[u];
          minHeap.emplace(dist[v], v);
        } else if (d + w == dist[v]) {
          ways[v] += ways[u];
          ways[v] %= kMod;
        }
    }

    return ways[dst];
  }
};

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class Solution {
  public int countPaths(int n, int[][] roads) {
    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];
      final int v = road[1];
      final int w = road[2];
      graph[u].add(new Pair<>(v, w));
      graph[v].add(new Pair<>(u, w));
    }

    return dijkstra(graph, 0, n - 1);
  }

  private int dijkstra(List<Pair<Integer, Integer>>[] graph, int src, int dst) {
    final int kMod = 1_000_000_007;
    long[] ways = new long[graph.length];
    Arrays.fill(ways, 0);
    long[] dist = new long[graph.length];
    Arrays.fill(dist, Long.MAX_VALUE);

    ways[src] = 1;
    dist[src] = 0;
    Queue<Pair<Long, Integer>> minHeap = new PriorityQueue<>(Comparator.comparing(Pair::getKey)) {
      { offer(new Pair<>(dist[src], src)); }
    };

    while (!minHeap.isEmpty()) {
      final long 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 < dist[v]) {
          dist[v] = d + w;
          ways[v] = ways[u];
          minHeap.offer(new Pair<>(dist[v], v));
        } else if (d + w == dist[v]) {
          ways[v] += ways[u];
          ways[v] %= kMod;
        }
      }
    }

    return (int) ways[dst];
  }
}

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

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

    return self._dijkstra(graph, 0, n - 1)

  def _dijkstra(
      self,
      graph: list[list[tuple[int, int]]],
      src: int,
      dst: int,
  ) -> int:
    kMod = 10**9 + 7
    ways = [0] * len(graph)
    dist = [math.inf] * len(graph)

    ways[src] = 1
    dist[src] = 0
    minHeap = [(dist[src], src)]

    while minHeap:
      d, u = heapq.heappop(minHeap)
      if d > dist[u]:
        continue
      for v, w in graph[u]:
        if d + w < dist[v]:
          dist[v] = d + w
          ways[v] = ways[u]
          heapq.heappush(minHeap, (dist[v], v))
        elif d + w == dist[v]:
          ways[v] += ways[u]
          ways[v] %= kMod

    return ways[dst]