寻找最短路径Dijkstra算法

 1 /**
 2      * 1.对于T中的每个顶点u，找到u的具有最小权重的连接边。所有到u的连接边都存储在queues.get(u)中。queues.get(u).peek()返回拥有最小权值
 3      * 的连接边。如果e.v已经在T中，将e从queues.get(u)中删除。
 4      * 2.比较所有这些边，并且找到那个具有cost[u]+e.getWeight()最小值的边。
 5      */
 6     public ShortestPathTree getShortestPath(int sourceIndex){
 7         List<Integer> T = new ArrayList<Integer>();
 8         T.add(sourceIndex);
 9
10         int numberOfVertices = vertices.size();
11
12         int[] parent = new int[numberOfVertices];
13         parent[sourceIndex] = -1;
14
15         int[] costs = new int[numberOfVertices];
16         for (int i = 0; i < costs.length; i++) {
17             costs[i] = Integer.MAX_VALUE;
18         }
19         costs[sourceIndex] = 0;
20
21         List<PriorityQueue<WeightedEdge>> queues = deepClone(this.queues);
22
23         while (T.size() < numberOfVertices) {
24             int v = -1;
25             int smallestCost = Integer.MAX_VALUE;
26             for (int u:T) {
27                 while(!queues.get(u).isEmpty() &&
28                         T.contains(queues.get(u).peek().v))
29                 {
30                     queues.get(u).remove();
31                 }
32                 if(queues.get(u).isEmpty()){
33                     continue;
34                 }
35                 WeightedEdge e = queues.get(u).peek();
36                 if(costs[u] + e.weight < smallestCost){
37                     v = e.v;
38                     smallestCost = costs[u] + e.weight;
39                     parent[v] = u;
40                 }
41             }
42             T.add(v);
43             costs[v] = smallestCost;
44         }
45         return new ShortestPathTree(sourceIndex, parent, T, costs);
46     }