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| 1 | +class Solution { |
| 2 | + public double maxProbability(int n, int[][] edges, double[] succProb, int start, int end) { |
| 3 | + // create the graph |
| 4 | + List<double[]>[] graph = new LinkedList[n]; |
| 5 | + for (int i = 0; i < n; i++) { |
| 6 | + graph[i] = new LinkedList<>(); |
| 7 | + } |
| 8 | + |
| 9 | + for (int i = 0; i < edges.length; i++) { |
| 10 | + double from = edges[i][0]; |
| 11 | + double to = edges[i][1]; |
| 12 | + double weight = succProb[i]; |
| 13 | + double[] m = new double[2]; |
| 14 | + m[0] = to; |
| 15 | + m[1] = weight; |
| 16 | + graph[edges[i][0]].add(m); |
| 17 | + double[] k = new double[2]; |
| 18 | + k[0] = from; |
| 19 | + k[1] = weight; |
| 20 | + graph[edges[i][1]].add(k); |
| 21 | + } |
| 22 | + |
| 23 | + // call dijkstra and return |
| 24 | + return dijkstra(start, end, graph); |
| 25 | + } |
| 26 | + |
| 27 | + class State{ |
| 28 | + int id; |
| 29 | + double proToStart; |
| 30 | + |
| 31 | + public State(int id, double proToStart){ |
| 32 | + this.id = id; |
| 33 | + this.proToStart = proToStart; |
| 34 | + } |
| 35 | + } |
| 36 | + |
| 37 | + private double dijkstra(int start, int end, List<double[]>[] graph){ |
| 38 | + double[] proTo = new double[graph.length]; |
| 39 | + // 初始化为一个去不到的值 |
| 40 | + Arrays.fill(proTo, -1); |
| 41 | + proTo[start] = 1; |
| 42 | + |
| 43 | + PriorityQueue<State> pq = new PriorityQueue<State>((a, b) -> { |
| 44 | + return Double.compare(b.proToStart, a.proToStart); |
| 45 | + }); |
| 46 | + pq.offer(new State(start, 1)); |
| 47 | + |
| 48 | + while (!pq.isEmpty()){ |
| 49 | + State cur = pq.poll(); |
| 50 | + int curid = cur.id; |
| 51 | + double curproToStart = cur.proToStart; |
| 52 | + |
| 53 | + if (curid == end) { |
| 54 | + return curproToStart; |
| 55 | + } |
| 56 | + |
| 57 | + if (proTo[curid] > curproToStart) { |
| 58 | + continue; |
| 59 | + } |
| 60 | + |
| 61 | + List<double[]> nexts = graph[curid]; |
| 62 | + for (double[] next: nexts) { |
| 63 | + double proToNext = proTo[curid] * next[1]; |
| 64 | + int idx = (int) next[0]; |
| 65 | + if (proToNext > proTo[idx]) { |
| 66 | + proTo[idx] = proToNext; |
| 67 | + pq.offer(new State(idx, proToNext)); |
| 68 | + } |
| 69 | + } |
| 70 | + } |
| 71 | + |
| 72 | + return 0; |
| 73 | + } |
| 74 | +} |
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