1

我正在尝试将新创建的对象添加到网格和图形中。具体来说，如何使用 for 循环有效地将节点添加到图中。它们的网格被设置为双数组，以更新初始视图。（对象网格是模型并通过推送更新视图）。我还使用 for 循环中 i 和 j 中的项目设置了一个 HashMap，以定义对象的键。但是，为了建立一个图形以便以后计算从节点到节点的最短路径，我需要将这些节点添加到图形中。我想使用 Djikstra 的算法来计算最短路径。我可以创建一个条件语句来定义网格中的角节点和边缘节点是什么，并定义哪些项目将具有双向边缘，但这似乎是一个“长切”。

下面是关于我如何创建前两项（创建双数组和 HashMap）的构造函数代码：

// **Constructor
// Construct a new Grid object. Sets up a HashMap of square object in order efficiently to get 
// and add Square objects later.
public ObjectGrid(Integer width, Integer height) 
{
    // View
    gui = new GUI();        // Instantiate GUI
    boardView = new BoardView(width,height);        // Instantiate BoardView

    // Initialize Gui, set time and add simulation event listener to model (ObjectGrid)
    gui.initGUI(BoardView);
    gui.addSimulationEventListener(this);

    // Initialize turnInSteps variable count to 0
    turnInSteps = 0;

    // Initialize numberOfDays variable count to 0
    numberOfDays = 1;

    // Instantiate HashMap
    objectHashMap = new HashMap();

    // Declare new object grid using double array of type objects.
    // Size determined by parameter Integer values in constructor
    myObjectGrid = new ObjectType[width][height];

    // Instantiate Graph object
    Graph graph = new ListGraph();

    // For loop sets up the initial grid with ObejctType using a double array. After
    // the completion of this loop, the grid will have XXX objects in the grid
    // each with a reference to an object. Objects are also added 
    // to HashMap using coordinates as keys and square objects as values.
    for(int i = 0; i < width; i++) 
    {  
        // Iterate through rows
        for(int j = 0; j < height; j++) 
        {  
            // Iterate through columns
            myObjectGrid[i][j] = new ObjectType();  // Instantiate a new Square at each row/column using default constructor
            gridView.addObjectView(myObjectGrid[i][j].getObjectView(), i, j);  // Add object view to each row/column placement
            String hashMapKey = (i + ", " + j); // Use values from i and j as Key for objects HashMap
            myObjectGrid[i][j].setID(hashMapKey);  // Add ID's for each ObjectView to display in object using values from i and j
            objectHashMap.add(hashMapKey, myObjectGrid[i][j]);  // Add object to HashMap using string value of coordinates as key
            listGraph.add(myObjectGrid[i][j]);

            // Pseudo code
            if (i != (width-height) && (j != height) etc) 
            {
                listGraph.addBidirectionalEdge(mySquareGrid[i][j], (mySquareGrid[i][j+1]), 1);
                listGraph.addBidirectionalEdge(mySquareGrid[i][j], (mySquareGrid[i+1][j]), 1);
                listGraph.addBidirectionalEdge(mySquareGrid[i][j], (mySquareGrid[i+1][j+1]), 1);
            }
        }
    }
}
4

1 回答 1

0

通过我所做的单元测试，图中的顶点返回 null。

原因是以下几行：

listGraph.addBidirectionalEdge(mySquareGrid[i][j], (mySquareGrid[i][j+1]), 1);
listGraph.addBidirectionalEdge(mySquareGrid[i][j], (mySquareGrid[i+1][j]), 1);
listGraph.addBidirectionalEdge(mySquareGrid[i][j], (mySquareGrid[i+1][j+1]), 1);

您正在初始化mySquareGrid并使用来自未来的节点创建边。j+1这些带有和的节点i+1此时为空。

尝试最简单的解决方案 - 在网格初始化之后，通过网格执行相同的循环并创建图形的边缘。

此外，您可以尝试将其更改为以下内容：

listGraph.addBidirectionalEdge(mySquareGrid[i][j-1], (mySquareGrid[i][j]), 1);
listGraph.addBidirectionalEdge(mySquareGrid[i-1][j], (mySquareGrid[i][j]), 1);
listGraph.addBidirectionalEdge(mySquareGrid[i-1][j-1], (mySquareGrid[i][j]), 1);
于 2012-12-24T05:03:15.190 回答