寻找最短路径Dijkstra算法

/**
     * 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
     * 的连接边。如果e.v已经在T中，将e从queues.get(u)中删除。
     * 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。
     */
    public ShortestPathTree getShortestPath(int sourceIndex){
        List<Integer> T = new ArrayList<Integer>();
        T.add(sourceIndex);
        
        int numberOfVertices = vertices.size();
        
        int[] parent = new int[numberOfVertices];
        parent[sourceIndex] = -1;
        
        int[] costs = new int[numberOfVertices];
        for (int i = 0; i < costs.length; i++) {
            costs[i] = Integer.MAX_VALUE;
        }
        costs[sourceIndex] = 0;
        
        List<PriorityQueue<WeightedEdge>> queues = deepClone(this.queues);
        
        while (T.size() < numberOfVertices) {
            int v = -1;
            int smallestCost = Integer.MAX_VALUE;
            for (int u:T) {
                while(!queues.get(u).isEmpty() &&
                        T.contains(queues.get(u).peek().v))
                {
                    queues.get(u).remove();
                }
                if(queues.get(u).isEmpty()){
                    continue;
                }
                WeightedEdge e = queues.get(u).peek();
                if(costs[u] + e.weight < smallestCost){
                    v = e.v;
                    smallestCost = costs[u] + e.weight;
                    parent[v] = u;
                }
            }
            T.add(v);
            costs[v] = smallestCost;
        }
        return new ShortestPathTree(sourceIndex, parent, T, costs);
    }