0

我创建了一个函数来计算两条线段的交点。

不幸的是,如果其中一个段是垂直的,则下面的代码将不起作用

    public static Point intersection(Segment s1, Segment s2) {
    double x1 = s1.getP1().getX();
    double y1 = s1.getP1().getY() ;
    double x2 = s1.getP2().getX();
    double y2 = s1.getP2().getY() ;
    double x3 = s2.getP1().getX();
    double y3 = s2.getP1().getY();
    double x4 = s2.getP2().getX();
    double y4 = s2.getP2().getY();

    double d = (x1 - x2) * (y3 - y4) - (y1 - y2) * (x3 - x4);
    if (d == 0) {
        return null;
    }
    double xi = ((x3 - x4) * (x1 * y2 - y1 * x2) - (x1 - x2) * (x3 * y4 - y3 * x4)) / d;
    double yi = ((y3 - y4) * (x1 * y2 - y1 * x2) - (y1 - y2) * (x3 * y4 - y3 * x4)) / d;
    Point p = new Point(xi, yi);
    if (xi < Math.min(x1, x2) || xi > Math.max(x1, x2)) {
        return null;
    }
    if (xi < Math.min(x3, x4) || xi > Math.max(x3, x4)) {
        return null;
    }
    return p;
}

当我有一条垂直线段时的问题,这个公式

double d = (x1 - x2) * (y3 - y4) - (y1 - y2) * (x3 - x4);

等于 0 并且该方法返回 null。

我该如何处理这个异常。

谢谢

4

3 回答 3

2
于 2015-04-24T22:25:44.453 回答
0

这是我的答案。我已经通过创建一个循环来测试它是否准确,该循环检查它给出的答案是否与 Boost 几何库给出的答案相同,并且他们同意每个测试,尽管我在下面写的那个比它快得多Boost中的那个。对于所有可能的线段对,该测试生成每个可能的线段,其中 x 是 [-3,2] 中的整数,y 是 [-3,2] 中的整数。

下面的代码认为在端点处相交的线段是相交的。T 形交叉口也被认为是交叉口。代码是 c++,但很容易适应任何语言。它基于不同的 stackoverflow 答案,但该答案没有正确处理端点。

它使用叉积方法,可以报告一个点是在给定射线的左边还是右边。

在数学上需要进行一些优化,但是这样做并没有显示出编译时g++ -O2的性能提升,有时甚至会降低性能!编译器能够进行这些优化,所以我更愿意让代码保持可读性。

// is_left(): tests if a point is Left|On|Right of an infinite line.
//    Input:  three points p0, p1, and p2
//    Return: >0 for p2 left of the line through p0 and p1
//            =0 for p2 on the line
//            <0 for p2 right of the line
//    See: Algorithm 1 "Area of Triangles and Polygons"
//    This is p0p1 cross p0p2.
extern inline coordinate_type_fp is_left(point_type_fp p0, point_type_fp p1, point_type_fp p2) {
  return ((p1.x() - p0.x()) * (p2.y() - p0.y()) -
          (p2.x() - p0.x()) * (p1.y() - p0.y()));
}

// Is x between a and b, where a can be lesser or greater than b.  If
// x == a or x == b, also returns true. */
extern inline coordinate_type_fp is_between(coordinate_type_fp a,
                                            coordinate_type_fp x,
                                            coordinate_type_fp b) {
  return x == a || x == b || (a-x>0) == (x-b>0);
}

// https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
extern inline bool is_intersecting(const point_type_fp& p0, const point_type_fp& p1,
                                   const point_type_fp& p2, const point_type_fp& p3) {
  const coordinate_type_fp left012 = is_left(p0, p1, p2);
  const coordinate_type_fp left013 = is_left(p0, p1, p3);
  const coordinate_type_fp left230 = is_left(p2, p3, p0);
  const coordinate_type_fp left231 = is_left(p2, p3, p1);

  if (p0 != p1) {
    if (left012 == 0) {
      if (is_between(p0.x(), p2.x(), p1.x()) &&
          is_between(p0.y(), p2.y(), p1.y())) {
        return true; // p2 is on the line p0 to p1
      }
    }
    if (left013 == 0) {
      if (is_between(p0.x(), p3.x(), p1.x()) &&
          is_between(p0.y(), p3.y(), p1.y())) {
        return true; // p3 is on the line p0 to p1
      }
    }
  }
  if (p2 != p3) {
    if (left230 == 0) {
      if (is_between(p2.x(), p0.x(), p3.x()) &&
          is_between(p2.y(), p0.y(), p3.y())) {
        return true; // p0 is on the line p2 to p3
      }
    }
    if (left231 == 0) {
      if (is_between(p2.x(), p1.x(), p3.x()) &&
          is_between(p2.y(), p1.y(), p3.y())) {
        return true; // p1 is on the line p2 to p3
      }
    }
  }
  if ((left012 > 0) == (left013 > 0) ||
      (left230 > 0) == (left231 > 0)) {
    if (p1 == p2) {
      return true;
    }
    return false;
  } else {
    return true;
  }
}

测试代码:

BOOST_AUTO_TEST_CASE(small, *boost::unit_test::disabled()) {
  for (double x0 = -3; x0 < 3; x0++) {
    for (double y0 = -3; y0 < 3; y0++) {
      for (double x1 = -3; x1 < 3; x1++) {
        for (double y1 = -3; y1 < 3; y1++) {
          for (double x2 = -3; x2 < 3; x2++) {
            for (double y2 = -3; y2 < 3; y2++) {
              for (double x3 = -3; x3 < 3; x3++) {
                for (double y3 = -3; y3 < 3; y3++) {
                  point_type_fp p0{x0, y0};
                  point_type_fp p1{x1, y1};
                  point_type_fp p2{x2, y2};
                  point_type_fp p3{x3, y3};
                  linestring_type_fp ls0{p0,p1};
                  linestring_type_fp ls1{p2,p3};
                  BOOST_TEST_INFO("intersection: " << bg::wkt(ls0) << " " << bg::wkt(ls1));
                  BOOST_CHECK_EQUAL(
                      path_finding::is_intersecting(p0, p1, p2, p3),
                      bg::intersects(ls0, ls1));
                }
              }
            }
          }
        }
      }
    }
  }
}
于 2021-02-12T21:09:21.810 回答
0

我已经尝试在没有测试的情况下对其进行编码......我希望它有效!^^

public static Point intersection(Segment s1, Segment s2) {
    // Components of the first segment's rect.
    Point v1 = new Point(s1.p2.x - s1.p1.x, s1.p2.y - s1.p1.y); // Directional vector
    double a1 = v.y;
    double b1 = -v.x;
    double c1 = v1.x * s1.p1.y - s1.v.y * s1.p1.x;

    // Components of the second segment's rect.
    Point v2 = new Point(s2.p2.x - s2.p1.x, s2.p2.y - s2.p1.y);
    double a2 = v2.y;
    double b2 = -v2.x;
    double c2 = v2.x * s2.p1.y - s2.v.y * s2.p1.x;

    // Calc intersection between RECTS.
    Point intersection = null;
    double det = a1 * b2 - b1 * a2;
    if (det != 0) {
        intersection = new Point(
            (b2 * (-c1) - b1 * (-c2)) / det;
            (a1 * (-c2) - a2 * (-c1)) / det;
        );
    }

    // Checks ff segments collides.
    if (
        intersection != null &&
        (
            (s1.p1.x <= intersection.x && intersection.x <= s1.p2.x) ||
            (s1.p2.x <= intersection.x && intersection.x <= s1.p1.x)
        ) &&
        (
            (s1.p1.y <= intersection.y && intersection.y <= s1.p2.y) ||
            (s1.p2.y <= intersection.y && intersection.y <= s1.p1.y)
        ) &&
        (
            (s2.p1.x <= intersection.x && intersection.x <= s2.p2.x) ||
            (s2.p2.x <= intersection.x && intersection.x <= s2.p1.x)
        ) &&
        (
            (s2.p1.y <= intersection.y && intersection.y <= s2.p2.y) ||
            (s2.p2.y <= intersection.y && intersection.y <= s2.p1.y)
        )
    )
        return intersection;

    return null;
};
于 2016-05-13T17:59:54.520 回答