3112. 访问消失节点的最少时间

访问消失节点的最少时间

解法一: Dijkstra 算法

java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
class Solution {
private static class Node {
private int id;
private int distance;

private static Node of(int id, int distance) {
Node node = new Node();
node.id = id;
node.distance = distance;
return node;
}
}

public int[] minimumTime(int n, int[][] edges, int[] disappear) {
List<List<Node>> grid = new ArrayList<>(n);
for (int i = 0; i < n; i++) {
grid.add(new ArrayList<>());
}
for (int[] edge : edges) {
grid.get(edge[0]).add(Node.of(edge[1], edge[2]));
grid.get(edge[1]).add(Node.of(edge[0], edge[2]));
}
int[] ans = new int[n];
Arrays.fill(ans, -1);
ans[0] = 0;
PriorityQueue<Node> priorityQueue = new PriorityQueue<>((node1, node2) -> node1.distance - node2.distance);
priorityQueue.add(Node.of(0, 0));
while (!priorityQueue.isEmpty()) {
Node node = priorityQueue.poll();
if (node.distance > ans[node.id]) {
continue;
}
for (Node nextNode : grid.get(node.id)) {
if (node.distance + nextNode.distance < disappear[nextNode.id] &&
(ans[nextNode.id] == -1 || node.distance + nextNode.distance < ans[nextNode.id])) {
ans[nextNode.id] = node.distance + nextNode.distance;
priorityQueue.offer(Node.of(nextNode.id, ans[nextNode.id]));
}
}
}
return ans;
}
}
作者

wuhunyu

发布于

2024-07-18

更新于

2024-09-05

许可协议