743. Network Delay Time

El texto de la tarea se traduce del ruso para el idioma seleccionado. El código no cambia.

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.

Ejemplo:

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

Output: 2

C# solución

coincidente/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++ solución

borrador automático, revisar antes de enviar
#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 solución

coincidente/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 solución

coincidente/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 solución

coincidente/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 solución

coincidente/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 grafo 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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