743. Network Delay Time
선택한 UI 언어에 맞게 문제 텍스트를 러시아어에서 번역합니다. 코드는 변경하지 않습니다.
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.
예제:
Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2
C# 해법
매칭됨/원본using 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++ 해법
자동 초안, 제출 전 검토#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 해법
매칭됨/원본import 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 해법
매칭됨/원본var 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 해법
매칭됨/원본import 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 -1Go 해법
매칭됨/원본package 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 그래프 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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