Codeforces 32E hide-and-seek evaluate 2 points about the ry of the mirror reflection Calculation

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It should be noted that the mirror is not used as an obstacle when it is collocated with two people, but it is consistent with the wall in other situations.

# Include <cstdio> # include <iostream> # include <algorithm> # include <string. h> # include <math. h> using namespace STD; # define point pointconst double EPS = 1e-8; const double Pi = ACOs (-1.0); double ABS (Double X) {return x> 0? X:-X;} int SGN (Double X) {If (FABS (x) <EPS) return 0; If (x <0) Return-1; else return 1 ;} struct point {Double X, Y; void put () {printf ("(%. 0lf, %. 0lf) \ n ", x, y);} Point () {} Point (double _ x, double _ y) {x = _ x; y = _ y ;} point operator-(const point & B) const {return point (X-B. x, Y-B. y);} // Cross Product double operator ^ (const point & B) const {return x * B. y-y * B. x;} // dot product double operator * (const point & B) const {return x * B. X + Y * B. Y ;}/ /Rotate the angle B (radian value) around the origin, and change void transxy (double B) after X and Y {double Tx = x, Ty = y; X = TX * Cos (B)-ty * sin (B); y = TX * sin (B) + ty * Cos (B );}}; struct line {point S, E; void put () {S. put (); E. put ();} line () {} line (point _ s, point _ e) {S = _ s; E = _ E ;} // calculates the intersection of two straight lines. // if the first value is 0, the line is overlapped. If the value is 1, the line is parallel. If the value is 0, the line is intersecting, 2 is the intersection. // The intersection is meaningful when the first value is 2. Pair <int, point> operator & (const line & B) const {point res = s; if (SGN (S-e) ^ (B. s-b.e) = 0) {If (SGN (s-b.e) ^ (B. s-b.e) = 0) return make_pair (0, Res); // coincidence else return make_pair (1, Res); // parallel} double T = (s-b.s) ^ (B. s-b.e)/(S-e) ^ (B. s-b.e); Res. X + = (E. x-s.x) * t; Res. Y + = (E. y-s.y) * t; return make_pair (2, Res) ;}}; double dist (point a, point B) {return SQRT (a-B) * (a-B);} // * determines the intersection of line segments bool inter (line L1, line l2) {returnmax (l1.s. x, l1.e. x)> = min (l2.s. x, l2.e. x) & MAX (l2.s. x, l2.e. x)> = min (l1.s. x, l1.e. x) & MAX (l1.s. y, l1.e. y)> = min (l2.s. y, l 2. e. y) & MAX (l2.s. y, l2.e. y)> = min (l1.s. y, l1.e. y) & SGN (l2.s-l1.e) ^ (l1.s-l1.e) * SGN (l2.e-l1.e) ^ (l1.s-l1.e )) <= 0 & SGN (l1.s-l2.e) ^ (l2.s-l2.e) * SGN (l1.e-l2.e) ^ (l2.s-l2.e) <= 0 ;} point into ric_point (point P1, point L1, point L2) {point ret; If (ABS (l1.x-l2.x) <EPS) {ret. y = p1.y; ret. X = 2 * l1.x-p1.x; return ret;} If (ABS (l1.y-l2.y) <EPS) {ret. X = p1.x; ret. y = 2 * l1.y-p1.y; return ret;} If (l1.x> l2.x -EPS & l1.x <l2.x + EPS) {ret. X = (2 * l1.x-p1.x); ret. y = p1.y;} else {Double K = (l1.y-l2.y)/(l1.x-l2.x); ret. X = (2 * K * l1.x + 2 * K * p1.y-2 * K * l1.y-K * p1.x + p1.x)/(1 + K * k ); ret. y = p1.y-(Ret. x-p1.x)/K;} return ret;} bool Gongxian (point a, point B, point C) {return ABS (. y-b.y) * (. x-c.x)-(. y-c.y) * (. x-b.x) <EPS;} Point A, B; line PEO, Wal, Mir; bool work () {// PEO. put (); WAL. put (); If (Gongxian (A, Mir. s, Mir. e) & Gongxian (B, Mir. s, Mir. e) {//. put (); B. put (); Mir. put (); puts ("collocated"); return! Inter (Wal, PEO);} else if (Inter (MIR, PEO) return false; If (! Inter (Wal, PEO) return true; // puts ("Gege"); If (Inter (Wal, Mir) return false; point c = effecric_point (B, Mir. s, Mir. e); point D = pairic_point (A, Mir. s, Mir. e); // C. put (); D. put (); line AC, BD; AC. S = A, AC. E = C; BD. S = B, Bd. E = D; If (! Inter (AC, Mir) return false; If (! Inter (BD, Mir) return false; If (Inter (AC, wal) return false; If (Inter (BD, wal) return false; return true ;} int main () {While (~ Scanf ("% lf", &. x, &. y) {scanf ("% lf", & B. x, & B. y); scanf ("% lf", & Wal. s. x, & Wal. s. y, & Wal. e. x, & Wal. e. y); scanf ("% lf", & Mir. s. x, & Mir. s. y, & Mir. e. x, & Mir. e. y); PEO. S = A, PEO. E = B; Work ()? Puts ("yes"): puts ("no");} return 0 ;}