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
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;
}
}
- 理解 Dijkstra's algorithm 如何尋找最小花費路徑
- 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