车位iou计算

车位检测中,判断多帧图像检测出的车位是否是同一个车位.计算其IOU.

判断一个点是否在一个四边形内

Approach : Let the coordinates of four corners be A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4). And coordinates of the given point P be (x, y)

1. Calculate area of the given rectangle, i.e., area of the rectangle ABCD as area of triangle ABC + area of triangle ACD.
   Area A = [ x1(y2 – y3) + x2(y3 – y1) + x3(y1-y2)]/2 + [ x1(y4 – y3) + x4(y3 – y1) + x3(y1-y4)]/2
2. Calculate area of the triangle PAB as A1.
3. Calculate area of the triangle PBC as A2.
4. Calculate area of the triangle PCD as A3.
5. Calculate area of the triangle PAD as A4.
6. If P lies inside the triangle, then A1 + A2 + A3 + A4 must be equal to A.

#include  
using namespace std; 
  
/* A utility function to calculate area of  
   triangle formed by (x1, y1), (x2, y2) and 
  (x3, y3) */
float area(int x1, int y1, int x2, int y2, 
                            int x3, int y3) 
{ 
    return abs((x1 * (y2 - y3) + x2 * (y3 - y1) +  
                x3 * (y1 - y2)) / 2.0); 
} 
  
/* A function to check whether point P(x, y)  
   lies inside the rectangle formed by A(x1, y1),  
   B(x2, y2), C(x3, y3) and D(x4, y4) */
bool check(int x1, int y1, int x2, int y2, int x3,  
             int y3, int x4, int y4, int x, int y) 
{ 
    /* Calculate area of rectangle ABCD */
    float A = area(x1, y1, x2, y2, x3, y3) +  
              area(x1, y1, x4, y4, x3, y3); 
  
    /* Calculate area of triangle PAB */
    float A1 = area(x, y, x1, y1, x2, y2); 
  
    /* Calculate area of triangle PBC */
    float A2 = area(x, y, x2, y2, x3, y3); 
  
    /* Calculate area of triangle PCD */
    float A3 = area(x, y, x3, y3, x4, y4); 
  
    /* Calculate area of triangle PAD */
    float A4 = area(x, y, x1, y1, x4, y4); 
  
    /* Check if sum of A1, A2, A3 and A4  
       is same as A */
    return (A == A1 + A2 + A3 + A4); 
} 
  
/* Driver program to test above function */
int main() 
{ 
    /* Let us check whether the point P(10, 15) 
      lies inside the rectangle formed by A(0, 10), 
      B(10, 0) C(0, -10) D(-10, 0) */
    if (check(0, 10, 10, 0, 0, -10, -10, 0, 10, 15)) 
        cout << "yes"; 
    else
        cout << "no"; 
    return 0; 
}