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| 1 | +/** |
| 2 | + * Title: Reachable Nodes in Subdivided Graph |
| 3 | + * Description: You are given an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge. |
| 4 | + * Author: Hasibul Islam |
| 5 | + * Date: 04/05/2023 |
| 6 | + */ |
| 7 | + |
| 8 | +/** |
| 9 | + * @param {number[][]} edges |
| 10 | + * @param {number} maxMoves |
| 11 | + * @param {number} n |
| 12 | + * @return {number} |
| 13 | + */ |
| 14 | + |
| 15 | +const initializeGraph = (n) => { |
| 16 | + let G = []; |
| 17 | + for (let i = 0; i < n; i++) { |
| 18 | + G.push([]); |
| 19 | + } |
| 20 | + return G; |
| 21 | +}; |
| 22 | +const addEdgeWithCostToG = (G, Edges) => { |
| 23 | + for (const [u, v, cost] of Edges) { |
| 24 | + G[u].push([v, cost]); |
| 25 | + G[v].push([u, cost]); |
| 26 | + } |
| 27 | +}; |
| 28 | + |
| 29 | +const reachableNodes = (edges, maxMoves, n) => { |
| 30 | + let g = initializeGraph(n); |
| 31 | + addEdgeWithCostToG(g, edges); |
| 32 | + return dijkstra(g, n, edges, maxMoves); |
| 33 | +}; |
| 34 | + |
| 35 | +const MIN = Number.MIN_SAFE_INTEGER; |
| 36 | +const dijkstra = (g, n, edges, maxMoves) => { |
| 37 | + let pq = new MaxPriorityQueue({ priority: (x) => x[0] }); |
| 38 | + let blood = Array(n).fill(MIN); // for each node, save maximum moves left to reach each node |
| 39 | + pq.enqueue([maxMoves, 0]); |
| 40 | + let res = 0; |
| 41 | + while (pq.size()) { |
| 42 | + let [hp, cur] = pq.dequeue().element; |
| 43 | + if (blood[cur] != MIN) continue; |
| 44 | + blood[cur] = hp; |
| 45 | + res++; |
| 46 | + for (const [next_node, cost] of g[cur]) { |
| 47 | + let next_hp = hp - cost - 1; // cut the cost for move to adjacent child |
| 48 | + if (next_hp < 0 || blood[next_node] != MIN) continue; |
| 49 | + pq.enqueue([next_hp, next_node]); |
| 50 | + } |
| 51 | + } |
| 52 | + for (const [u, v, cost] of edges) { |
| 53 | + let uv = blood[u] == MIN ? 0 : blood[u]; |
| 54 | + let vu = blood[v] == MIN ? 0 : blood[v]; |
| 55 | + res += Math.min(cost, uv + vu); |
| 56 | + } |
| 57 | + return res; |
| 58 | +}; |
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