You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.

We will send a signal from a given node k. Return the minimum time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.

Example 1:

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

Example 2:

Input: times = [[1,2,1]], n = 2, k = 1 Output: 1 

Example 3:

Input: times = [[1,2,1]], n = 2, k = 2 Output: -1 

Constraints:

• 1 <= k <= n <= 100
• 1 <= times.length <= 6000
• times[i].length == 3
• 1 <= ui, vi <= n
• ui != vi
• 0 <= wi <= 100
• All the pairs (ui, vi) are unique. (i.e., no multiple edges.)

題目給定一個矩陣 times, 一個整數 n, 還有一個整數 k

其中每個 entry, times[i] = [$u_i, v_i, w_i]$ 代表 從在一條從 $u_i$ 到 $v_i$ 的路徑 且花費是 $w_i$

n 代表具有 label 1 到 n 個 vertex

k 代表從 label k 的 vertex 出發

要求寫一個演算法來找出從 k 發送封包:

如果可以傳送封包到所有 vertex的話, 封包傳達到所有 vertex 所需花費的最少時間

如果不能傳送到所有 vertex 則回傳 -1

這題的困難點在於如何找到最小花費路徑

首先需要透過 times 矩陣來建立 adjacencyList

然後 透過 Dijkstra's algorithm

因為封包可以同時運發

所以直覺的作法是對 adjacencyList 做 BFS

透過 minHeap 找出當下鄰近的 vertex 中篩選出要找最小花費的邊

並且把累計 weight 的最大值紀錄下來

為了避免重複走，要有一個 HashSet visit 來紀錄走訪過的 vertex

假設走完所有 adjacencyList 的 vertex

假設 visit 的長度 是 n 代表可以走完所有 vertex

否則就是沒有辦法走完

import java.util.*; public class Solution { static class AdjacentNode { final private int Weight; final private int Node; public AdjacentNode(int weight, int node) { this.Weight = weight; this.Node = node; } } public int networkDelayTime(int[][] times, int n, int k) { HashSet<Integer> visit = new HashSet<>(); HashMap<Integer, List<AdjacentNode>> adjacentMap = new HashMap<>(); for (int[] t : times) { int source = t[0]; int target = t[1]; int weight = t[2]; if (!adjacentMap.containsKey(source)) { adjacentMap.put(source, new ArrayList<>()); } adjacentMap.get(source).add(new AdjacentNode(weight, target)); } // Dijkstra Algorithm PriorityQueue<AdjacentNode> queue = new PriorityQueue<>(Comparator.comparingInt(a -> a.Weight)); // add init k queue.add(new AdjacentNode(0, k)); int time = 0; while(queue.size() != 0) { AdjacentNode node = queue.poll(); if (node != null) { if (visit.contains(node.Node)) { continue; } visit.add(node.Node); time = Math.max(time, node.Weight); List<AdjacentNode> adjList = adjacentMap.get(node.Node); if (adjList != null) { for (AdjacentNode adjNode : adjList) { if (!visit.contains(adjNode.Node)) { queue.add(new AdjacentNode(node.Weight + adjNode.Weight, adjNode.Node)); } } } } } if (visit.size() == n) { return time; } return -1; } }

1. 理解 Dijkstra's algorithm 如何尋找最小花費路徑
2. 理解 MinHeap

• 透過給定的 times 來建立 adjacencyList
• 透過 Dijkstra's algorithm 來找尋花費 最少的路徑
• 透過 MinHeap 與 BFS 來實作 Dijkstra's algorithm