Question

下图说明了创建Manhattan图时遇到的问题：

该框围绕着与现有线重叠的大部分线[（tx，midy） - （sx，midy）]（在下面的代码中由psegment表示）。我已经删除了重叠的箭头（和尾部），并且对如何检查重叠感到有些困惑。

以下是有问题的代码：

  Line2D.Double segment = new Line2D.Double( sx, midy, tx, midy );

  // Associate the middle-y point with the bounds of the target object.
  // On subsequent draws of targets with a similar mid-y, make sure that
  // there are no overlapping lines.
  //
  if( midPointMap.put( midy, segment ) != null ) {
    //if( midy == 90 ) {
    // New Line.
    //
    System.err.printf( "NEW: (%3.2f, %3d)-(%3.2f, %3d)\n", sx, midy, tx,
                       midy );

    for( Line2D.Double psegment : midPointMap.getValues( midy ) ) {
      // Previous Line.
      //
      System.err.printf( "OLD: (%3.2f, %3d)-(%3.2f, %3d)\n",
                         psegment.getX1(), midy, psegment.getX2(), midy );
    }
    //}
  }

  // Line for the bus.
  //
  result.moveTo( sx, midy );
  result.lineTo( tx, midy );

这是另一个示例图片，让您了解曼哈顿布局：

在上面的图中，Dialog和Window之间的线重叠（在此缩放时不太明显）。该图像说明了如何存在多个子类，因此检测重叠必须考虑沿同一y线的多个源（sx，sy）的多个目标（tx，ty）。

midPointMap变量是一个哈希集，每个键可以包含多个值：

  private MultiValueMap<Integer, Line2D.Double> midPointMap =
    new MultiValueMap<Integer, Line2D.Double>();

这会将mid-y值映射到一组线段。

如果与现有线段重叠，如何不绘制线条？

更新＃1

请注意，每个“总线”的线段没有特定的顺序。

Answer 1

我可能会遗漏某些东西（例如，我不明白你为什么要这样做 - 也许你正在用不同颜色或其他东西画画？如果你只是想优化一些写操作，我我不确定你是否会真正获得任何收益。

但是，假设有充分的理由这样做，我认为以下算法可行：

确定所有水平线段，并按y位置降序和段长度降序排序
绘制第一个线段
将第二个线段的y位置与列表中所有具有相同y位置的前一行（在这种情况下，只是第一个）进行比较。如果您没有得到精确的y位置匹配，请绘制该段，然后对后续段重复步骤3
如果确实得到了精确的y位置匹配，请比较较短段的终点，看它的x位置是否在较长段的两个终点的x位置之间。如果是，那么你有重叠。如果没有，请检查另一个终点。

我假设片段的布局是这样的，你不能有两个部分相互重叠的片段，像这样（片段是aA和bB）：

一个===== b ===甲=========乙

如果你确实有这种可能性，那么你将不得不决定如何解决这个问题。

PS - 请务必添加为什么的简要说明，以消除这些细分。我很好奇！

Answer 2

这是一个完整的解决方案：

  // If line segments would overlap, this gets set to false.
  //
  boolean drawSegment = true;

  Line2D.Double segment = new Line2D.Double( sx, midy, tx, midy );

  // Associate the middle-y point with the bounds of the target object.
  // On subsequent draws of targets with a similar mid-y, make sure that
  // there are no overlapping lines.
  //
  if( midPointMap.put( midy, segment ) != null ) {
    // Check previous lines for overlap. Each previous line segment has
    // values in the form: (sx, mid-y)-(tx, mid-y), which map to
    // (getX1(), midy)-(getX2(), midy).
    //
    for( Line2D.Double psegment : midPointMap.getValues( midy ) ) {
      // If the lines have the same source point, and differ in their
      // target point, then they might overlap
      //
      if( sx == psegment.getX1() && tx != psegment.getX2() ) {
        double pdx = psegment.getX1() - psegment.getX2();
        double cdx = sx - tx;

        // At this juncture: the mid-y points are the same, the source
        // points of the previous segment and the current segment are the
        // same, and the target points of the segments differ.
        //
        // If both lines go in the same direction (relative to the same
        // source point), then they overlap. The difference of the tx
        // and X2 points is how much overlap exists.
        //
        // There are two actionable possibilities: (1) psegment is longer
        // than the current segment; or (2) psegment is shorter.
        //
        // If psegment is longer, then no segment must be drawn. If
        // psegment is shorter, the difference between psegment and the
        // current segment must be drawn.
        //
        if( tx < sx && psegment.getX2() < sx ) {
          // SEGMENT IS TO THE LEFT OF SOURCE
          //
          if( pdx > cdx ) {
            // If the previous segment is longer, then draw nothing.
            //
            drawSegment = false;
          }
          else {
            // If the previous segment is shorter, then draw the
            // difference. That is, change the source point for
            // this segment to the target point of the previous segment.
            //
            sx = psegment.getX2();
          }
        }
        else if( tx > sx && psegment.getX2() > sx ) {
          // SEGMENT IS TO THE RIGHT OF SOURCE
          //
          if( pdx < cdx ) {
            // If the previous segment is longer, then draw nothing.
            //
            drawSegment = false;
          }
          else {
            // If the previous segment is shorter, then draw the
            // difference. That is, change the source point for
            // this segment to the target point of the previous segment.
            //
            sx = psegment.getX2();
          }
        }
      }
    }
  }

  // Draw the line for the bus.
  //
  if( drawSegment ) {
    result.moveTo( sx, midy );
    result.lineTo( tx, midy );
  }

如果可以优化（或简化），我真的很想知道。

重叠的线段

2 个答案: