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  • js迪杰斯特拉算法求最短路径

     

    1.后台生成矩阵

    名词解释和下图参考:https://blog.csdn.net/csdnxcn/article/details/80057574

    double[,] arr = new double[allVertices.Count(), allVertices.Count()]; //矩阵 

    //allVertices所有三维坐标点的集合

    //lines 所有两点的连线

    for (int i = 0; i < allVertices.Count(); i++)
    {
    for (int j = 0; j < allVertices.Count(); j++)
    {
    var start1 = allVertices[i].Point; //起点
    var end1 = allVertices[j].Point; //终点
    //lines 两点的连线集合
    var line = lines.FirstOrDefault(ee => (ee.StartPoint == start1 && ee.EndPoint == end1)|| (ee.StartPoint == end1 && ee.EndPoint == start1/*起点终点互换*/));
    if (start1 == end1)
    {//同一个点
    arr[i, j] = 0;
    }
    else
    {
    if (line != null)
    {
    arr[i, j] = double.Parse(line.Remark); //长度
    }
    else
    {//两点未连接 此路不通
    arr[i, j] =1.0/0.0; //Infinity
    }
    }
    }
    }

    return arr;

    2.dijkstra算法

    /**
    * Dijkstra算法
    *
    * @author wupanpan@baidu.com
    * @date 2014-03-26
    */

    /**
    * @const
    */
    var POS_INFINITY = Infinity;

    /**
    * @param {number} sourceV 源点的索引,从0开始
    * @param {Array} adjMatrix 图的邻接矩阵,是一个二维数组
    */

    function dijkstra(sourceV, adjMatrix) {
    var set = [],
    path = [],

    dist = [];
    distCopy = [],
    vertexNum = adjMatrix.length;

    var temp, u,
    count = 0;

    // 初始化
    for (var i = 0; i < vertexNum; i++) {
    distCopy[i] = dist[i] = POS_INFINITY;
    set[i] = false;
    }
    distCopy[sourceV] = dist[sourceV] = 0;

    while (count < vertexNum) {
    u = distCopy.indexOf(Math.min.apply(Math, distCopy));
    set[u] = true;
    distCopy[u] = POS_INFINITY;

    for (var i = 0; i < vertexNum; i++) {
    if (!set[i] && ((temp = dist[u] + adjMatrix[u][i]) < dist[i])) {
    distCopy[i] = dist[i] = temp;
    path[i] = u;
    }
    }
    count++;
    }

    return {
    path: path,
    dist: dist
    };
    }

    /**
    * @param {number} v 源点索引, 从0开始
    * @param {number} d 非源点索引, 从0开始
    * @param {Array} adjMatrix 图的邻接矩阵,是一个二维数组
    */
    function searchPath(v, d, adjMatrix) {
    var graph = dijkstra(v, adjMatrix),
    path = graph.path,
    dist = graph.dist;

    var prev = path[d],
    queue = [],
    str = '';

    queue.push(d);
    while(prev != v) {
    queue.push(prev);
    prev = path[prev];
    }
    queue.push(v);

    for (var j = queue.length - 1; j >= 0; j--) {
    str +=queue.pop() + '->';
    }
    console.log('path',str);
    var arr=str.split('->');
    if(str.endsWith('->')){
    arr.pop();
    }
    var rarr=[];//字符串数组转int数组
    for(var i=0;i<arr.length;i++){
    rarr.push(parseInt(arr[i]));
    }
    return rarr;
    }


    /**
    * 测试数据
    */
    var adjM = [
    [0, 4, 2, POS_INFINITY, POS_INFINITY, POS_INFINITY],
    [4, 0, 1, 5, POS_INFINITY, POS_INFINITY],
    [2, 1, 0, 8, 10, POS_INFINITY],
    [POS_INFINITY, 5, 8, 0, 2, 6],
    [POS_INFINITY, POS_INFINITY, 10, 2, 0, 3],
    [POS_INFINITY, POS_INFINITY, POS_INFINITY, 6, 3, 0]
    ];

    3.使用算法求最短路径

    5个点坐标如上图 虚线表示两点相连

    1:  0,0,0
    2:  1,1,0
    3:  -1,-1,0
    4:  2,0,0
    5:  0,-1,0

    请求后台生成的矩阵为:

    var pathMatrix = [
    [
    0,
    1.73,
    1.73,
    "Infinity",
    1
    ],
    [
    1.73,
    0,
    "Infinity",
    1.73,
    "Infinity"
    ],
    [
    1.73,
    "Infinity",
    0,
    "Infinity",
    "Infinity"
    ],
    [
    "Infinity",
    1.73,
    "Infinity",
    0,
    2.23
    ],
    [
    1,
    "Infinity",
    "Infinity",
    2.23,
    0
    ]
    ];

    var ret = searchPath(4, 1, pathMatrix); //从第5点到第2点的最短路径
    console.log('index', ret);

     (索引从0开始,对应到图上是 5->1->2)

     4.使用threejs画出路径

    (黑色连线;  红绿蓝为xyz辅助线)

    geometryPoint = new THREE.BoxGeometry(0.2, 0.2, 0.2);
    var materialPoint = new THREE.MeshBasicMaterial({
    color: 0xff00ff,
    side: THREE.DoubleSide
    });
    circlePoint1 = new THREE.Mesh(geometryPoint, materialPoint);
    circlePoint1.position.set(0, 0, 0);
    scene.add(circlePoint1);

    circlePoint2 = circlePoint1.clone();
    circlePoint2.position.set(1, 1, 0);
    scene.add(circlePoint2);

    circlePoint3 = circlePoint1.clone();
    circlePoint3.position.set(-1, 1, 0);
    scene.add(circlePoint3);


    circlePoint4 = circlePoint1.clone();
    circlePoint4.position.set(2, 0, 0);
    scene.add(circlePoint4);


    circlePoint5 = circlePoint1.clone();
    circlePoint5.position.set(0, -1, 0);
    scene.add(circlePoint5);

    scene.add(new THREE.AxesHelper(300));

    //画路径

    var ret = searchPath(4, 1, pathMatrix);   //从第5点到第2点的最短路径
    console.log('index', ret);

    var geometry1 = new THREE.Geometry();
    for (var i = 0; i < ret.length; i++) {
    console.log("circlePoint" + (ret[i] + 1));
    var pointObj = eval("circlePoint" + (ret[i] + 1));
    console.log('position', pointObj.position);
    geometry1.vertices.push(pointObj.position);
    }
    var line = new THREE.Line(geometry1, new THREE.LineBasicMaterial({
    color: 'black'
    }), THREE.LinePieces);
    scene.add(line);

    //补充

    //threejs求三维两点的距离

    var distance = circlePoint4.position.distanceTo(circlePoint5.position);
    console.log(distance);

    From:https://www.cnblogs.com/xuejianxiyang/p/9776319.html

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  • 原文地址:https://www.cnblogs.com/xuejianxiyang/p/9776319.html
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