Question

我正试图了解大学的TSP计划，我不得不承认我发现它非常困难。

基本上我有一个Lat值数组和一个Lng值数组，使用hasrsine，我已将距离转换为矩阵。
现在从哪里开始？
我必须找到最短的距离，访问所有80个城镇。我想到最好的方法来做到这一点。我应该制作最近邻算法吗？我应该只存储排列和总距离。或者，还有更好的方法？我不得不承认，我认为这对一年级学生来说有点困难！无论如何这里是我的代码，它很糟糕，小程序等给了我们。

public class Brain {

    //These are the names of the 80 towns and their north and west GPS co-ordinates

    static double north[] = {53.855,52.794,54.350,53.433,52.992,54.117,53.328,54.800,54.863,55.071,54.502,54.343,51.746,54.660,51.680,54.597,53.091,53.175,55.136,52.831,53.976,53.944,53.861,53.991,51.622,52.354,51.897,54.996,54.322,53.714,53.348,54.009,54.500,52.085,53.345,52.846,52.502,54.345,53.272,52.677,53.728,53.106,52.648,52.059,51.708,53.783,54.851,54.957,55.053,52.665,52.447,53.727,53.197,51.904,54.750,52.131,53.382,52.266,54.248,53.116,53.522,52.863,52.396,54.210,52.451,54.590,53.633,52.714,54.267,53.245,54.830,52.679,52.474,52.268,53.515,53.267,52.257,53.800,52.334,51.952};
    static double west[] = {-6.538,-6.165,-6.655,-7.950,-6.987,-9.167,-8.219,-7.790,-6.284,-6.508,-8.190,-6.260,-8.735,-5.670,-9.453,-5.930,-7.913,-6.525,-7.456,-6.932,-6.719,-8.095,-9.299,-7.360,-8.886,-7.712,-8.470,-7.307,-5.703,-6.350,-6.260,-6.405,-6.770,-7.640,-7.051,-8.981,-6.566,-7.640,-9.049,-6.292,-6.878,-6.065,-7.256,-9.507,-8.531,-8.917,-5.811,-7.720,-6.946,-8.624,-9.486,-7.800,-8.567,-8.957,-6.610,-8.642,-6.591,-8.270,-6.971,-7.324,-7.338,-8.200,-6.945,-5.882,-9.055,-7.290,-8.183,-8.869,-8.483,-9.306,-7.470,-7.814,-8.162,-9.696,-8.851,-7.500,-7.129,-9.533,-6.458,-7.846};
    String names[] = {"Ardee","Arklow","Armagh","Athlone","Athy","Ballina","Ballinasloe","Ballybofe","Ballymena","Ballymoney","Ballyshannon","Banbridge","Bandon","Bangor","Bantry","Belfast","Birr","Blessington","Buncrana","Carlow","Carrickmacross","Carrick-On-Shannon","Castlebar","Cavan","Clonakilty","Clonmel","Cork","Derry","Downpatrick","Drogheda","Dublin","Dundalk","Dungannon","Dungarvan","Edenderry","Ennis","Enniscorthy","Enniskillen","Galway","Gorey","Kells","Kilcoole","Kilkenny","Killarney","Kinsale","Knock","Larne","Letterkenny","Limavady","Limerick","Listowel","Longford","Loughrea","Macroom","Magherafelt","Mallow","Maynooth","Mitchelstown","Monaghan","Mountmellick","Mullingar","Nenagh","New-Ross","Newcastle","Newcastle-West","Omagh","Roscommon","Shannon","Sligo","Spiddal","Strabane","Thurles","Tipperary","Tralee","Tuam","Tullamore","Waterford","Westport","Wexford","Youghal"};
    static double[][] matrix = new double[80][80];
    boolean visit[]=new boolean[80];
    boolean valid = true;

    public static void fillmatrix (){
        double tote = 0;
        for(int i=1;i<80;i++){
            for(int j=1;j<80;j++){
                matrix[i][j]=getDistance(north[i],west[j],north[j],west[j]);
            }
        }
    }

    public String compute () {          
        String solution ="";
        for (int i=0;i<80;i++){
            solution+=(char)(i+40);
        }
        solution+="Anything you add on after the 80 characters will be " + 
            "printed in the textbox (e.g. you can compute the distance)";
        return solution;    
    }

    public static double getDistance(double lat1, double lon1, double lat2, double lon2){
        double R = 6371;
        double dLat = Math.toRadians((lat2-lat1));
        double dLon = Math.toRadians((lon2-lon1)); 
        double a = Math.sin(dLat/2) * Math.sin(dLat/2) + 
            Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) *
            Math.sin(dLon/2) * Math.sin(dLon/2); 
        double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
        double d = R * c;         
        return d;
    }    
}

Answer 1

除了最微小的情况之外，检查所有排列对于任何事情来说都是非常昂贵的。

幸运的是，TSP有一大堆着名的启发式算法。 Take your pick

Answer 2

解决最近邻启发式问题的最佳方法是将算法分解为子问题。

选择一个随机城市。
找到最近的未访问城市，然后去那里。
是否有任何无人值守的城市？如是，重复步骤2.
返回第一个城市。

接下来，解决每个子问题。

在名称数组中选择一个随机值
通过遍历北西阵列并计算距离，检查所有城市与随机城市的距离。将访问过的城市存储在一个阵列中。
重复2次
返回第一个城市

然后，考虑伪代码并尝试编写Java代码。

Answer 3

关于SourceCode

首先，我可能会建议您将城市坐标的定义更改为城市类，并且，由于您预先定义了城市，您可能希望将其“导出”到外部文件中并将其加载到文件中beggining。例如，

public class City{
    public double North;
    public double West;
    public String Name;
    public double getDistanceToCity(City target){
        double R = 6371;
        double dLat = Math.toRadians((target.North-this.North));
        double dLon = Math.toRadians((target.West-this.West)); 
        double a = Math.sin(dLat/2) * Math.sin(dLat/2) + 
            Math.cos(Math.toRadians(this.North)) * Math.cos(Math.toRadians(this.West)) *
            Math.sin(dLon/2) * Math.sin(dLon/2); 
        double d = R * 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));         
        return d;}}

然后，例如，在将您的文件读入“ArrayList Cities”之后，您将通过以下方式对您的矩阵进行打包：

double[][] distances = new double[Cities.size()][Cities.size()];
int i=0;
int j=0;
for(City start:Cities){
    for(City end:Cities){
        distances[i][j]=start.getDistanceToCity(end);
        j++;}
    i++;}

获得相同的矩阵。在这种情况下，您可以更改输入数据大小，例如测试您的算法与10个城市的正确性，并且在所有工作完成后运行80.此外，在这种情况下，您将亲眼看到这个问题的硬度 - 简单在40和41城市运行......

接下来，关于算法......

这是NP难题，所以你的排列会很长时间。我会建议Held-Karp算法，这个算法要快得多，但很难实现，需要大量内存。（例如，建议的方法if（n * n * 2 ^ n），对于n = 80是10 ^ 27，并且排列是n！，其中10 ^ 121对于n = 80 - 我指的是解决任务所需的操作）

无论如何，现在最好的解决方案是线性编程（不记得确切的方法，但是......）。你可以找一些考虑使用图表方法的算法。

创建旅行推销员计划，排列是一种很好的技巧吗？

3 个答案: