Calculate the geometric poj 2826 an easy problem ?! (Determine the line segment position and calculate the intersection point)

This question is so sad. I have been wa for a long time because I have misunderstood the question.

Let's talk about how to solve the problem first:

1. When the two line segments do not want to be handed over, the result is 0.00. Parallel conditions are excluded here;

2. When either of the two line segments is parallel to the X axis, the result is also 0.00;

3. When the two line segments intersect but are in the same line (determined in parallel), the result is also 0.00;

4. Another situation is that, if one of the above boards blocks the following boards, the same result would be 0.00;

In addition to the above four cases, we can use the cross point obtained by the cross product, and then obtain the Y value of the horizontal plane that can be filled with water above the intersection, and use the cross product to calculate the area.

This question has been wa for a long time because I think that when two boards don't want to be handed in, I can use the ground as a bottom, and then ask for the area of the trapezoid. As a result, WA has been a long time, so sad ......

The following code is for reference only ......

# Include <cstring> # include <cstdlib> # include <cstdio> # include <iostream> # include <cmath> using namespace STD; const double EPS = 1e-8; int dblcmp (Double X) {If (FABS (x) <EPS) return 0; return x> 0.0? 1:-1;} typedef struct point {Double X, Y; point () {} Point (double Tx, double ty) {x = Tx, y = ty ;}} point; double dotmult (point a, point B, point C) {return (C. x-a.x) * (B. x-c.x) + (C. y-a.y) * (B. y-c.y);} double xmult (point P1, point P2, point P0) {return (p1.x-Snapshot X) * (p2.y-Snapshot y)-(p1.y-Snapshot y) * (p2.x-0000x);} int iscross (point a, point B, point C, point D) // The normalized intersection {int T1 = dblcmp (xmult (C,, b); int t2 = dblcmp (xmult (D, a, B); int T3 = Dblcmp (xmult (A, C, D); int t4 = dblcmp (xmult (B, c, d )); if (T1 * t2 <0 & T3 * t4 <0) return 1; return 0;} int iscross2 (point a, point B, point C, point D) // non-normalized intersection {int T1 = dblcmp (xmult (C, A, B); int t2 = dblcmp (xmult (D, a, B )); int T3 = dblcmp (xmult (A, C, D); int t4 = dblcmp (xmult (B, c, d )); if (T1 * t2 <0 & T3 * t4 <0) return 1; if (T1 = 0 & dblcmp (dotmult (a, B, c)> = 0) return 1; if (t2 = 0 & dblcmp (dotmult (a, B, d)> = 0) return 1; if (t3 = 0 & DB Lcmp (dotmult (c, d, A)> = 0) return 1; if (t4 = 0 & dblcmp (dotmult (c, d, B)> = 0) return 1; return 0;} Point getcrosspoint (point a, point B, double YY) {double dx = B. x-a.x; double DY = B. y-a.y; If (dblcmp (dx) = 0) {return point (. x, YY);} else {Double K = Dy/dx; If (dblcmp (K) = 0) {return point (max (. x, B. x), YY);} else {return point (yy-a.y)/K +. x, YY) ;}}int isupon (point, point); double solvetringle (Point P1, point P2, point P3, point P4) {If (dblcmp (p2.x-p1.x) * (p4.y-p3.y)-(p2.y-p1.y) * (p4.x-p3.x) = 0) return 0.0; If (isupon (P1, P2, P3, P4) return 0.00; double T1 = xmult (P1, P3, p4); double t2 = xmult (P2, P3, P4); point P0; struct x = (T1 * p2.x-T2 * p1.x)/(t1-t2 ); p0.y = (T1 * p2.y-T2 * p1.y)/(t1-t2); double YY1 = max (p1.y, p2.y); double yy2 = max (p3.y, p4.y); double YY; if (dblcmp (yy1-p0.y) = 0 | dblcmp (yy2-p0.y) = 0) return 0.00; If (dblcmp (yy1-yy2)> = 0) {YY = yy2; point P11 = getcrosspoint (P1, P2, YY); point P22 = getcrosspoint (P3, P4, YY ); return ABS (xmult (P11, P22, P0)/2.0;} else {YY = YY1; point P11 = getcrosspoint (P1, P2, YY ); point P22 = getcrosspoint (P3, P4, YY); Return ABS (xmult (P11, P22, P0)/2.0 ;}} void myswap (point & A, point & B) {point t; T. X =. x, t. y =. y;. X = B. x,. y = B. y; B. X = T. x, B. y = T. y;} int isupon (point P1, point P2, point P3, point P 4) // determine whether to block ...... {If (iscross2 (P1, P2, P4, point (p4.x, 10000.0) return 1; return 0 ;}int main () {// freopen ("poj2826in.txt ", "r", stdin); // freopen ("myout.txt", "W", stdout); int t; double ans; Point P1, P2, P3, P4; scanf ("% d", & T); While (t --) {scanf ("% lf", & p1.x, & p1.y, & p2.x, & p2.y); scanf ("% lf", & p3.x, & p3.y, & p4.x, & p4.y); If (dblcmp (p1.y-p2.y)> 0) myswap (P1, P2); If (dblcmp (p3.y-p4.y)> 0) myswap (P3, P4); If (dblcmp (p2.y-p4.y) <0) {myswap (P1, P3); myswap (P2, P4);} If (dblcmp (p2.y-p1.y) = 0) ans = 0.00; else if (dblcmp (p4.y-p3.y) = 0) ans = 0.00; else if (iscross2 (P1, P2, P3, P4) ans = solvetringle (P1, P2, P2, p3, P4); else ans = 0.00; printf ("%. 2f \ n ", ANS);} return 0 ;}