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.
Esempio:
Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2
C# soluzione
abbinato/originaleusing 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++ soluzione
bozza automatica, rivedere prima dell'invio#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 soluzione
abbinato/originaleimport 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 soluzione
abbinato/originalevar 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 soluzione
abbinato/originaleimport 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 soluzione
abbinato/originalepackage 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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