Java中使用Above below Primitive测试两条线是否相交

上方下方原语是通过比较它们的端点来检查线是否相交的方法。这是一个Java程序，它使用上面下面的原语来测试两条线是否相互交叉。如果线的一个端点在另一条线的左端点的左侧，并且其右端点在另一条线的端点的右侧，则此方法仅返回 true。

例子

Input:
      End Points of first line
      x1 = 2, y1 = 3
      x2 = 6, y2 = 4
      End Points of second line
      p1 = 3, q1 = 4
      p2 = 7, q2 = 1
Output:
      Equation 1: (1)x+(-4)y+(10) = 0
      Equation 2: (-3)x+(-4)y+(25) = 0
      Intersecting

方法

Find the segment first line as a1.
         if a1 < 0
             Find the segment of line 2 as a2
         if a2 >= 0
             print intersecting
         otherwise
             print not intersecting.
         else if a1 >0
             Find the segment of line 2 as a2
         if a2 <= 0
             print intersecting.
         otherwise
             print not intersecting.

下面是上述方法的实现。

Java

// This is a java program to find whether two lines
// intersect or not using above and below primitive
public class Main {
    public static void check(int x1, int x2, int y1, int y2,
                             int p1, int p2, int q1, int q2)
    { // Segment of line 1 is stored as a1
  
        int a1 = (y2 - y1) * p1 + (x1 - x2) * q1
                 + (x2 * y1 - x1 * y2);
  
        if (a1 < 0) {
            // Segment of line 2 is stored as a2
            int a2 = (y2 - y1) * p2 + (x1 - x2) * q2
                     + (x2 * y1 - x1 * y2);
  
            if (a2 >= 0)
                System.out.println("Intersecting");
  
            else if (a2 < 0)
                System.out.println("Not intersecting");
        }
  
        else if (a1 > 0) {
  
            int a2 = (y2 - y1) * p2 + (x1 - x2) * q2
                     + (x2 * y1 - x1 * y2);
  
            if (a2 <= 0)
                System.out.println("Intersecting");
  
            else if (a2 > 0)
                System.out.println("Not intersecting");
        }
  
        // lines are coincinding
        else
            System.out.println(
                "points are lying on the line");
    }
    // Driver Code
    public static void main(String args[])
    {
        // Taking the coordinates of first line as input
        int x1 = 2, y1 = 3;
        int x2 = 6, y2 = 4;
  
        // Equation of line using slop point form
        System.out.println("Equation 1: (" + (y2 - y1)
                           + ")x+(" + (x1 - x2) + ")y+("
                           + (x2 * y1 - x1 * y2) + ") = 0");
  
        // Taking the coordinates of second line as input
        int p1 = 3, q1 = 4;
        int p2 = 7, q2 = 1;
  
        // Equation of line using slop point form
        System.out.println("Equation 2: (" + (q2 - q1)
                           + ")x+(" + (p1 - p2) + ")y+("
                           + (p2 * q1 - p1 * q2) + ") = 0");
  
        check(x1, x2, y1, y2, p1, p2, q1, q2);
    }
}

输出

Equation 1: (1)x+(-4)y+(10) = 0
Equation 2: (-3)x+(-4)y+(25) = 0
Intersecting

时间复杂度： O(1)