POJ 2826 An Easy Problem ?! Good question

POJ 2826 An Easy Problem ?! Good question

The general idea is that two blocks form a slot, asking how much rain can be installed in the slot, and paying attention to the vertical falling of rain, the idea is also very simple, that is, classification discussion is a bit bad.
1. If the two line segments do not overlap or are parallel, set 0;
2. There is a parallel X axis with 0 loaded;
3. If it overwrites the following, install 0;
4. For others, cross product is used to calculate the area.

Directly run the Code:

# Include
  
   
# Include
   
    
# Include
    
     
Using namespace std; const double eps = 1e-8; struct point {double x; double y ;}; struct line {point a; point B ;}l1, l2; double ans; // calculate the Cross Product double xmult (point p0, point p1, point p2) {return (p1.x-converted X) * (p2.y-converted y)-(p2.x-converted X) * (p1.y-policy);} int dblcmp (double n) {if (fabs (n)
     
      
0? 1:-1;} // determines whether or not the intersection exists. How to intersection int judge (line l1, line l2) {double d1 = dblcmp (max (l1.a. x, l1. B. x)-min (l2.a. x, l2. B. x); double d2 = dblcmp (max (l2.a. x, l2. B. x)-min (l1.a. x, l1. B. x); double d3 = dblcmp (max (l1.a. y, l1. B. y)-min (l2.a. y, l2. B. y); double d4 = dblcmp (max (l2.a. y, l2. B. y)-min (l1.a. y, l1. B. y); double d5 = dblcmp (xmult (l2.a, l1.a, l1. B); double d6 = dblcmp (xmult (l2. B, l1.a, l1. B )); double d7 = dblcmp (xmult (l1.a, l2.a, l2. B); double d8 = dblcmp (xmult (l1. B, l2.a, l2. B )); if (d1> = 0 & d2> = 0 & d3> = 0 & d4> = 0) {if (d5 * d6> 0 | d7 * d8> 0) return 0; // non-Intersecting else if (d5 = 0 & d6 = 0) return 1; // if (d5 = 0 | d6 = 0 | d7 = 0 | d8 = 0) return 2; // endpoint intersection else return 3; // normalized intersection} return 0;} // returns the slope bool getslope (line l, double & k) {double t = l. a. x-l. B .x; if (t = 0) return false; k = (l. a. y-l. B .y)/t; return true;} // calculates the intersection of line segments point getIntersect (line l1, line l2) {point p; double A1 = l1. B. y-l1.a. y; double B1 = l1.a. x-l1. B. x; double C1 = (l1. B. x-l1.a. x) * l1.a. y-(l1. B. y-l1.a. y) * l1.a. x; double A2 = l2. B. y-l2.a. y; double B2 = l2.a. x-l2. B. x; double C2 = (l2. B. x-l2.a. x) * l2.a. y-(l2. B. y-l2.a. y) * l2.a. x; p. x = (C2 * B1-C1 * B2)/(A1 * B2-A2 * B1); p. y = (C1 * A2-C2 * A1)/(A1 * B2-A2 * B1); if (p. x =-0) p. x = 0; return p;} // calculate the point of point getbiggerY (point a, point B) {point q; if (dblcmp (. y-b.y)> 0) {q. x =. x; q. y =. y;} else {q. x = B. x; q. y = B. y;} return q;} double getarea (point p, point p1, point p2) {point q; double a; if (dblcmp (p1.y-p2.y)> = 0) {q. y = p2.y; q. x = p. x + (p1.x-p. x) * (p2.y-p. y)/(p1.y-p. y); // obtain the coordinate a = fabs (xmult (p, p2, q)/2; // calculate the area} else {q. y = p1.y; q. x = p. x + (p2.x-p. x) * (p1.y-p. y)/(p2.y-p. y); a = fabs (xmult (p, p1, q)/2;} return a;} int main () {int t; cin> t; while (t --) {point p1, p2, p; cin> l1.a. x> l1.a. y> l1. B. x> l1. B. y; cin> l2.a. x> l2.a. y> l2. B. x> l2. B. y; if (judge (l1, l2) <= 1) // determines whether the intersection exists {ans = 0; cout <0 <
      
        0) {// when the slope symbols of the two line segments are the same, int d1 = dblcmp (K1-K2); int d2 = dblcmp (k2 ); if (d1> 0 & d2> 0 & dblcmp (p2.x-p1.x) * dblcmp (p2.x-p. x) <= 0 | d1 <0 & d2> 0 & dblcmp (p1.x-p2.x) * dblcmp (p1.x-p. x) <= 0 | d1> 0 & d2 <0 & dblcmp (p1.x-p2.x) * dblcmp (p1.x-p. x) <= 0 | d1 <0 & d2 <0 & dblcmp (p2.x-p1.x) * dblcmp (p2.x-p. x) <= 0) // coverage // {ans = 0; // cout <3 <