743. Network Delay Time
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.
Beispiel:
Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2
C# Lösung
zugeordnet/originalusing 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++ Lösung
Auto-Entwurf, vor dem Einreichen prüfen#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 Lösung
zugeordnet/originalimport 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 Lösung
zugeordnet/originalvar 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 Lösung
zugeordnet/originalimport 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 Lösung
zugeordnet/originalpackage 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 Graph 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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