|
1 | | -class minHeap { |
2 | | - constructor() { |
3 | | - this.heap = []; |
4 | | - this.heap.push([Number.MIN_SAFE_INTEGER, 0]); |
5 | | - } |
6 | | - insert([a, b]) { |
7 | | - this.heap.push([a, b]); |
8 | | - this.upheap(this.heap.length - 1); |
9 | | - } |
10 | | - upheap(pos) { |
11 | | - let tmp = this.heap[pos]; |
12 | | - while (tmp[1] < this.heap[parseInt(pos / 2)][1]) { |
13 | | - this.heap[pos] = this.heap[parseInt(pos / 2)]; |
14 | | - pos = parseInt(pos / 2); |
15 | | - } |
16 | | - this.heap[pos] = tmp; |
17 | | - } |
18 | | - get() { |
19 | | - if (this.heap.length === 2) { |
20 | | - return this.heap.pop(); |
21 | | - } |
22 | | - let res; |
23 | | - res = this.heap[1]; |
24 | | - this.heap[1] = this.heap.pop(); |
25 | | - this.downheap(1, this.heap.length - 1); |
26 | | - return res; |
27 | | - } |
28 | | - downheap(pos, len) { |
29 | | - let tmp, i; |
30 | | - tmp = this.heap[pos]; |
31 | | - while (pos <= parseInt(len / 2)) { |
32 | | - i = pos * 2; |
33 | | - if (i < len && this.heap[i][1] < this.heap[i + 1][1]) i++; |
34 | | - if (tmp[1] <= this.heap[i][1]) break; |
35 | | - this.heap[pos] = this.heap[i]; |
36 | | - pos = i; |
37 | | - } |
38 | | - this.heap[pos] = tmp; |
39 | | - } |
40 | | - size() { |
41 | | - return this.heap.length - 1; |
42 | | - } |
43 | | - top() { |
44 | | - return this.heap[1]; |
45 | | - } |
46 | | -} |
47 | | -function solution(N, road, K) { |
48 | | - let answer = 0; |
49 | | - let minH = new minHeap(); |
50 | | - let distance = Array(N + 1).fill(1e9); |
51 | | - let graph = Array(N + 1); |
52 | | - for (let i = 0; i < graph.length; i++) { |
53 | | - graph[i] = Array(); |
54 | | - } |
55 | | - |
56 | | - for (let [a, b, c] of road) { |
| 1 | +function solution(n, edges, k) { |
| 2 | + let answer; |
| 3 | + let graph = Array(n + 1); |
| 4 | + for (let i = 0; i < graph.length; i++) graph[i] = Array(); |
| 5 | + let check = Array(n + 1).fill(0); |
| 6 | + let dist = Array(n + 1).fill(1e9); |
| 7 | + edges.forEach(([a, b, c]) => { |
57 | 8 | graph[a].push([b, c]); |
58 | | - graph[b].push([a, c]); |
59 | | - } |
| 9 | + }); |
60 | 10 |
|
61 | | - minH.insert([1, 0]); |
62 | | - distance[1] = 0; |
63 | | - while (minH.size() > 0) { |
64 | | - let [now, time] = minH.get(); |
65 | | - if (time > distance[now]) continue; |
66 | | - for (let [next, nextTime] of graph[now]) { |
67 | | - if (time + nextTime < distance[next]) { |
68 | | - distance[next] = time + nextTime; |
69 | | - minH.insert([next, distance[next]]); |
70 | | - } |
| 11 | + dist[1] = 0; |
| 12 | + for (let i = 1; i <= n; i++) { |
| 13 | + let min = 0; |
| 14 | + |
| 15 | + // min을 이용해서 찾아야하는 노드를 min에 입력 |
| 16 | + for (let j = 1; j <= n; j++) { |
| 17 | + if (check[j] === 0 && dist[j] < dist[min]) min = j; |
71 | 18 | } |
72 | | - } |
73 | 19 |
|
74 | | - for (let x of distance) { |
75 | | - if (x <= K) answer++; |
| 20 | + // min과 연결된 노드들 탐색 |
| 21 | + for (let [next, cost] of graph[min]) { |
| 22 | + if (dist[min] + cost < dist[next]) { |
| 23 | + dist[next] = dist[min] + cost; |
| 24 | + } |
| 25 | + } |
| 26 | + check[min] = 1; |
76 | 27 | } |
77 | | - return answer; |
| 28 | + return dist[k]; |
78 | 29 | } |
79 | 30 |
|
80 | 31 | console.log( |
81 | 32 | solution( |
82 | | - 5, |
| 33 | + 6, |
83 | 34 | [ |
84 | | - [1, 2, 1], |
85 | | - [2, 3, 3], |
86 | | - [5, 2, 2], |
87 | | - [1, 4, 2], |
88 | | - [5, 3, 1], |
89 | | - [5, 4, 2], |
| 35 | + [1, 2, 12], |
| 36 | + [1, 3, 4], |
| 37 | + [2, 1, 2], |
| 38 | + [2, 3, 5], |
| 39 | + [2, 5, 5], |
| 40 | + [3, 4, 5], |
| 41 | + [4, 2, 2], |
| 42 | + [4, 5, 5], |
| 43 | + [6, 4, 5], |
90 | 44 | ], |
91 | | - 3 |
| 45 | + 5 |
92 | 46 | ) |
93 | | -); // 4 |
| 47 | +); |
| 48 | +// 14 |
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