swift 基本图的算法，带权值的BFS广度优先搜索

Question

在swift中，想存储一个图的结构，在segmentfault上看到过一个代码，可以实现BFS，如下：

enum VertexState: Int {
    case white = 0
    case gray = 1
    case black = 2
}

class Vertex {
    var parent: Vertex?
    var color: VertexState!
    var distance: Int
    var vIndex: Int

    init (_ index: Int) {
        self.vIndex = index
        self.parent = nil
        self.color = VertexState.white
        self.distance = 100000
    }
}

var queue: [Vertex] = []

class Adjacency {
    var adjacency: [Vertex] = []
}

class Graph {
    var vertexArray: [Vertex]!
    var adjacencyList: [Adjacency]!

    init (vertexArray: [Vertex], adjacencyList: [Adjacency]) {
        self.vertexArray = vertexArray
        self.adjacencyList = adjacencyList
    }
}

func BFS(inout graph: Graph, sourceVertex: Vertex) {

    // sourceVertex置灰入队列
    sourceVertex.color = VertexState.gray
    sourceVertex.distance = 0
    sourceVertex.parent = nil
    queue.append(sourceVertex)


    // 遍历
    while queue.count > 0 {
        var u = queue[0]
        queue.removeAtIndex(0)

        var adj = graph.adjacencyList[u.vIndex]
        for var i = 0; i < adj.adjacency.count; i++ {
            var v = adj.adjacency[i]
            if v.color == VertexState.white {
                v.color = VertexState.gray
                v.distance = u.distance + 1
                v.parent = u
                queue.append(v)
            }

            u.color = VertexState.black
        }
    }
}

func printPath(g: Graph, s: Vertex, v: Vertex) {
    if v.vIndex == s.vIndex {
        print("\(s.vIndex)\t")
    } else if v.parent == nil {
        print("no path from s to v exists")
    } else {
        printPath(g, s, v.parent!)
        print("\(v.vIndex)\t")
    }
}


var vArray: [Vertex] = []
var adjList: [Adjacency] = []


for var i = 0; i < 8; i++ {
    var vertex: Vertex = Vertex(i)
    vArray.append(vertex)
}



// TEST CASE

var adj0: Adjacency = Adjacency()
adj0.adjacency = [vArray[1], vArray[2]]

var adj1: Adjacency = Adjacency()
adj1.adjacency = [vArray[0], vArray[3], vArray[4]]

var adj2: Adjacency = Adjacency()
adj2.adjacency = [vArray[0], vArray[5]]

var adj3: Adjacency = Adjacency()
adj3.adjacency = [vArray[1], vArray[4], vArray[6]]

var adj4: Adjacency = Adjacency()
adj4.adjacency = [vArray[1], vArray[3], vArray[6], vArray[7]]

var adj5: Adjacency = Adjacency()
adj5.adjacency = [vArray[2]]

var adj6: Adjacency = Adjacency()
adj6.adjacency = [vArray[3], vArray[4], vArray[7]]

var adj7: Adjacency = Adjacency()
adj7.adjacency = [vArray[4], vArray[6]]

adjList = [adj0, adj1, adj2, adj3, adj4, adj5, adj6, adj7]

var g: Graph = Graph(vertexArray: vArray, adjacencyList: adjList)

BFS(&g, vArray[0])
printPath(g, vArray[0], vArray[2])

但是我现在要处理的图是带权值的，求帮忙解决一下这个问题