javascript实现kruskal算法

<script>
    //图的构建
    function vnode() {
        this.visited = 0;
        this.vertex = 0;
        this.arcs = new Array();
    }
    function G() {
        this.adjlist = new Array();
    }
    //0  1  2  3  4  5
    var a = [[0, 7, 8, 0, 0, 9], //
    [0, 0, 0, 2, 5, 0], //
    [0, 0, 0, 5, 0, 6], //
    [0, 0, 0, 0, 9, 0], //
    [0, 0, 0, 0, 0, 7], //
    [0, 0, 0, 0, 0, 0]]; //
    function creategraph() {
        var g = new G();
        for (i = 0; i < 6; i++) {
            g.adjlist[i] = new vnode();
            g.adjlist[i].vertex = i;
            g.adjlist[i].arcs = (function () {
                var b = new Array();
                for (j = 0; j < 6; j++)
                    if (a[i][j]) { b.push(new Number(j)); b[b.length - 1].weight = a[i][j]; }
                return b;
            })();
        }
        return g;
    }

    var s = new Array();
    var g = creategraph();
    var arcs = new Array();
    for (i = 0; i < 6; i++) {
        for (j = 0; j < g.adjlist[i].arcs.length; j++) {
            var b = {};
            b.a = i;
            b.b = g.adjlist[i].arcs[j];
            b.c = g.adjlist[i].arcs[j].weight;
            arcs.push(b);
        }

    }
    arcs.sort(function (a, b) {
        return a.c - b.c;
    })

    function kruskal() {
        for (i = 0; i < 6; i++) {
            s.push([i]);
            s[i].e = new Array();
        }
        var flag = 0;
        loop: for (i = 0; i < arcs.length; i++) {//判断一条边是否可以被选取
            if (flag) { i = 0; flag = 0; continue; }
            for (j = 0; j < s.length; j++) {//检查选取的边的两个端点是否同时落在以构建的树中
                for (k = 0; k < s[j].length; k++) {
                    if (s[j][k] == arcs[i].a) {//如果一个端点落在一棵树上，继续检查另一个端点是否落在同一棵树上

                        for (l = 0; l < s[j].length; l++) {
                            if (s[j][l] == arcs[i].b) {//如果两端点都落在同一棵树上，则继续取下一条边进行验证
                                continue loop;

                            }
                            if (l == (s[j].length - 1)) {//不落在同一棵树上，则把一个端点所在的树和另一个端点所在的树合并
                                for (m = 0; m < s.length; m++) {
                                    for (n = 0; n < s[m].length; n++) {

                                        if (s[m][n] == arcs[i].b) {//找到另一端点所在的树，进行合并
                                            alert(arcs[i].b + "---" + arcs[i].a);
                                            for (v = 0; v < s[m].length; v++) {
                                                s[j].push(s[m][v]);
                                            }
                                            flag = 1;
                                            s.splice(m, 1);
                                            continue loop;

                                        }

                                    }

                                }

                            }

                        }

                    }

                }

            }

        }
        alert(s.length);
    }
    kruskal();
</script>