Poj 3304 segments [calculation ry] [relationship between a straight line and a line segment]

Question link: http://poj.org/problem? Id = 3304

T case, each case contains n lines to determine whether a straight line exists, so that all lines can have common points in this line. If yes is output, otherwise, no is output.

The meaning of the question can be changed to: Can two vertices in 2 * n endpoints of n straight lines form a straight line to meet this condition, brute force enumeration is enough, note that the two points selected each time in the 2 * n vertices of the enumeration should be used to determine whether they are repeated.

# Include <iostream> # include <stdio. h> # include <math. h> # include <cstring> # include <stdlib. h> using namespace STD; # define EPS 1e-8 # define zero (x)> 0? (X):-(x) <EPS) // if it is 0, 1 is returned. Otherwise, 0 # DEFINE _ sign (x)> EPS? 1 :( (x) <-EPS? 2: 0) // negative 0 returns 2 0 1 struct point {Double X; Double Y; point (const double & x = 0, const double & Y = 0 ): x (x), y (y) {}void in () {scanf ("% lf", & X, & Y) ;} void out () const {printf ("%. 2f %. 2f \ n ", x, y) ;}; struct line {point a, B; line (const point & A = point (), const point & B = point (): A (a), B (B) {} void in () {. in (); B. in ();} void out () const {. out (); B. out () ;};// calculate the cross (P1-P0) x (P2-P0) Double xmult (Point P1, point P2, point P0) {// P0 is the center return (p1.x-1_x) * (p2.y-1_y)-(p2.x-1_x) * (p1.y-policy);} // returns 1int same_side (point P1, point P2, line L) {return xmult (L. a, P1, L. b) * xmult (L. a, P2, L. b)> EPS;} line L [105]; point P [500]; int N; int judge (line LL) {If (_ sign (LL. a. x-LL. B .x) = 0 & _ sign (LL. a. y-LL. B .y) = 0) return 0; For (INT I = 0; I <n; I ++) {If (same_side (L [I]. a, L [I]. b, LL) = 1) return 0;} return 1;} int Main () {int t; scanf ("% d", & T); While (t --) {scanf ("% d", & N ); for (INT I = 0; I <n * 2; I ++) P [I]. in (); For (INT I = 0; I <n; I ++) {L [I]. A = P [I * 2]; L [I]. B = P [I * 2 + 1];} int flag = 0; For (INT I = 0; I <n * 2-1; I ++) for (Int J = I + 1; j <n * 2; j ++) {line LL; ll. A = P [I]; ll. B = P [J]; If (Judge (LL) = 1) {flag = 1; break ;}} if (flag = 1) cout <"Yes! "<Endl; else cout <" No! "<Endl ;}}

Poj 3304 segments [calculation ry] [relationship between a straight line and a line segment]