Poj 3304 segments-intersection of straight lines and Line Segments

/* Question: Ask some line segments if there is such a line: all these line segments have a common point projected on the line. If there is such a point, make the vertical line of this line, this vertical line is crossed by all line segments because the projection of all line segments covers this point, because their projection has this common point, it is to find a straight line that passes through all line segments. So, enumerate the straight lines composed of any two points to see if they meet the requirements */# include <stdio. h> # include <math. h> const double zero = 1e-8; struct point {Double X, Y ;}; struct segment // line segment {Point P1, P2;} s [110]; point PP [210]; // All Village Points int t, n; int equal (point P1, point P2) // determine whether it is the same point {If (FABS (p1.x-p2.x) <= zero & FABS (p1.y-p2.y) <= zero) return 1; return 0;} doubl E cross (point P, point V1, point V2) // performs the cross product {return (v2.x-v1.x) * (p. y-v1.y)-(P. x-v1.x) * (v2.y-v1.y);} int scrossl (point S1, point S2, point L1, point L2) // determine whether a straight line passes through a line segment {If (Cross (S1, l1, L2) * Cross (S2, L1, L2) <= 0) return 1; return 0;} int main () {int I, J, K; scanf ("% d", & T); While (t --) {scanf ("% d", & N); for (I = 0; I <N; ++ I) {scanf ("% lf", & S [I]. p1.x, & S [I]. p1.y, & S [I]. p2.x, & S [I]. p2.y); PP [I * 2 + 0]. X = s [I]. p1.x; PP [I * 2 + 0]. y = s [I]. p1.y; PP [I * 2 + 1]. X = s [I]. p2.x; PP [I * 2 + 1]. y = s [I]. p2.y;} for (I = 0; I <2 * n-1; I ++) {for (j = I + 1; j <2 * n; ++ J) // enumeration point {If (! Equal (PP [I], PP [J]) // not the same point {for (k = 0; k <n; k ++) // test all side if (! Scrossl (s [K]. p1, s [K]. p2, PP [I], PP [J]) break; If (k = N) Break ;}} if (k = N) break ;} if (k = N) printf ("Yes! \ N "); else printf (" No! \ N ") ;}return 0 ;}