743. Network Delay Time

Le texte du problème est traduit du russe pour la langue sélectionnée. Le code reste inchangé.

Дана сеть из узлов, помеченных от 1 до n. Также given times - список времен прохождения сигнала в виде направленных ребер times[i] = (ui, vi, wi), где ui - исходный узел, vi - целевой узел, а wi - время прохождения сигнала от источника до цели. Мы пошлем сигнал из заданного узла k. return минимальное время, которое поit is required всем узлам, чтобы получить сигнал. Если все узлы не могут получить сигнал, return -1.

Exemple:

Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2

Output: 2

C# solution

correspondant/original
using System;
using System.Collections.Generic;
public class Solution {
    public int NetworkDelayTime(int[][] times, int n, int k) {
        var graph = new Dictionary<int, List<int[]>>();
        for (int i = 1; i <= n; i++) {
            graph[i] = new List<int[]>();
        }
        foreach (var time in times) {
            graph[time[0]].Add(new int[]{time[1], time[2]});
        }
        var minHeap = new SortedSet<(int time, int node)>(Comparer<(int, int)>.Create((a, b) => a.time != b.time ? a.time - b.time : a.node - b.node));
        minHeap.Add((0, k));
        var minTime = new Dictionary<int, int>();
        for (int i = 1; i <= n; i++) {
            minTime[i] = int.MaxValue;
        }
        minTime[k] = 0;
        while (minHeap.Count > 0) {
            var (time, node) = minHeap.Min;
            minHeap.Remove(minHeap.Min);
            foreach (var neighbor in graph[node]) {
                int newTime = time + neighbor[1];
                if (newTime < minTime[neighbor[0]]) {
                    minHeap.Remove((minTime[neighbor[0]], neighbor[0]));
                    minTime[neighbor[0]] = newTime;
                    minHeap.Add((newTime, neighbor[0]));
                }
            }
        }
        int maxTime = int.MinValue;
        foreach (var t in minTime.Values) {
            if (t == int.MaxValue) return -1;
            maxTime = Math.Max(maxTime, t);
        }
        return maxTime;
    }
}

C++ solution

brouillon automatique, à relire avant soumission
#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public int NetworkDelayTime(int[][] times, int n, int k) {
        var graph = new unordered_map<int, List<int[]>>();
        for (int i = 1; i <= n; i++) {
            graph[i] = new List<int[]>();
        }
        foreach (var time in times) {
            graph[time[0]].push_back(new int[]{time[1], time[2]});
        }
        var minHeap = new SortedSet<(int time, int node)>(Comparer<(int, int)>.Create((a, b) => a.time != b.time ? a.time - b.time : a.node - b.node));
        minHeap.push_back((0, k));
        var minTime = new unordered_map<int, int>();
        for (int i = 1; i <= n; i++) {
            minTime[i] = int.MaxValue;
        }
        minTime[k] = 0;
        while (minHeap.size() > 0) {
            var (time, node) = minHeap.Min;
            minHeap.Remove(minHeap.Min);
            foreach (var neighbor in graph[node]) {
                int newTime = time + neighbor[1];
                if (newTime < minTime[neighbor[0]]) {
                    minHeap.Remove((minTime[neighbor[0]], neighbor[0]));
                    minTime[neighbor[0]] = newTime;
                    minHeap.push_back((newTime, neighbor[0]));
                }
            }
        }
        int maxTime = int.MinValue;
        foreach (var t in minTime.Values) {
            if (t == int.MaxValue) return -1;
            maxTime = max(maxTime, t);
        }
        return maxTime;
    }
}

Java solution

correspondant/original
import java.util.*;

public class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        Map<Integer, List<int[]>> graph = new HashMap<>();
        for (int i = 1; i <= n; i++) {
            graph.put(i, new ArrayList<>());
        }
        for (int[] time : times) {
            graph.get(time[0]).add(new int[]{time[1], time[2]});
        }

        PriorityQueue<int[]> minHeap = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        minHeap.add(new int[]{0, k});
        Map<Integer, Integer> minTime = new HashMap<>();
        for (int i = 1; i <= n; i++) {
            minTime.put(i, Integer.MAX_VALUE);
        }
        minTime.put(k, 0);

        while (!minHeap.isEmpty()) {
            int[] top = minHeap.poll();
            int time = top[0];
            int node = top[1];
            for (int[] neighbor : graph.get(node)) {
                int newTime = time + neighbor[1];
                if (newTime < minTime.get(neighbor[0])) {
                    minTime.put(neighbor[0], newTime);
                    minHeap.add(new int[]{newTime, neighbor[0]});
                }
            }
        }

        int maxTime = Collections.max(minTime.values());
        return maxTime == Integer.MAX_VALUE ? -1 : maxTime;
    }
}

JavaScript solution

correspondant/original
var networkDelayTime = function(times, n, k) {
    const graph = Array.from({ length: n + 1 }, () => []);
    for (const [u, v, w] of times) {
        graph[u].push([v, w]);
    }

    const minHeap = [[0, k]];
    const minTime = Array(n + 1).fill(Infinity);
    minTime[k] = 0;

    while (minHeap.length) {
        minHeap.sort((a, b) => a[0] - b[0]);
        const [time, node] = minHeap.shift();
        for (const [neighbor, t] of graph[node]) {
            const newTime = time + t;
            if (newTime < minTime[neighbor]) {
                minTime[neighbor] = newTime;
                minHeap.push([newTime, neighbor]);
            }
        }
    }

    const maxTime = Math.max(...minTime.slice(1));
    return maxTime === Infinity ? -1 : maxTime;
};

Python solution

correspondant/original
import heapq

def networkDelayTime(times, n, k):
    graph = {i: [] for i in range(1, n + 1)}
    for u, v, w in times:
        graph[u].append((v, w))

    min_heap = [(0, k)]
    min_time = {i: float('inf') for i in range(1, n + 1)}
    min_time[k] = 0

    while min_heap:
        time, node = heapq.heappop(min_heap)
        for neighbor, t in graph[node]:
            new_time = time + t
            if new_time < min_time[neighbor]:
                min_time[neighbor] = new_time
                heapq.heappush(min_heap, (new_time, neighbor))

    max_time = max(min_time.values())
    return max_time if max_time < float('inf') else -1

Go solution

correspondant/original
package main

import (
  "container/heap"
  "math"
)

type Edge struct {
  to, weight int
}

type MinHeap [][]int

func (h MinHeap) Len() int            { return len(h) }
func (h MinHeap) Less(i, j int) bool  { return h[i][0] < h[j][0] }
func (h MinHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
func (h *MinHeap) Push(x interface{}) { *h = append(*h, x.([]int)) }
func (h *MinHeap) Pop() interface{} {
  old := *h
  n := len(old)
  x := old[n-1]
  *h = old[0 : n-1]
  return x
}

func networkDelayTime(times [][]int, n int, k int) int {
  graph := make(map[int][]Edge)
  for _, time := range times {
    graph[time[0]] = append(graph[time[0]], Edge{time[1], time[2]})
  }

  minHeap := &MinHeap{}
  heap.Init(minHeap)
  heap.Push(minHeap, []int{0, k})
  minTime := make(map[int]int)
  for i := 1; i <= n; i++ {
    minTime[i] = math.MaxInt32
  }
  minTime[k] = 0

  for minHeap.Len() > 0 {
    t := heap.Pop(minHeap).([]int)
    time, node := t[0], t[1]
    for _, edge := range graph[node] {
      newTime := time + edge.weight
      if newTime < minTime[edge.to] {
        minTime[edge.to] = newTime
        heap.Push(minHeap, []int{newTime, edge.to})
      }
    }
  }

  maxTime := 0
  for _, t := range minTime {
    if t == math.MaxInt32 {
      return -1
    }
    if t > maxTime {
      maxTime = t
    }
  }
  return maxTime
}

Algorithm

Представьте graphe в виде списка смежности.

Используйте Dijkstra's algorithm для нахождения кратчайших путей от узла k до всех других узлов.

find максимальное значение среди кратчайших путей к узлам. Если какой-либо узел недостижим, return -1.

😎

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