Poj 3384 feng shui-returns the inner side of a polygon from the left side of the polygon and returns the semi-plane intersection.

/* Poj 3384 feng shui-returns the inner side of a polygon to the left side and returns the semi-plane intersection. Then, the farthest point is obtained. */# include <stdio. h> # include <math. h ># include <algorithm> using namespace STD; const double EPS = 1e-8; struct point {Double X, Y; point () {} Point (double A, double B ): X (A), y (B) {}} Dian [400 + 10]; point Jiao [400 + 10]; struct line {point S, E; double angle ;} xian [200 + 10]; int N, Yong; Double R; bool mo_ee (Double X, Double Y) {double ret = x-y; If (Ret <0) ret =-ret; I F (Ret <EPS) return 1; return 0;} bool mo_gg (Double X, Double Y) {return x> Y + EPS ;} // x> Y bool mo_ll (Double X, Double Y) {return x <Y-EPS;} // x <y bool mo_ge (Double X, Double Y) {return x> Y-EPS;} // x> = y bool mo_le (Double X, Double Y) {return x <Y + EPS ;} // x <= y point mo_intersection (point u1, point U2, point V1, point V2) {point ret = U1; Double T = (u1.x-v1.x) * (v1.y-v2.y)-(u1.y-V1. Y) * (v1.x-v2.x)/(u1.x-u2.x) * (v1.y-v2.y)-(u1.y-u2.y) * (v1.x-v2.x); ret. X + = (u2.x-u1.x) * t; ret. Y + = (u2.y-u1.y) * t; return ret;} double mo_xmult (point P2, point P0, Point P1) // P1 returns a negative value on the left of P2, return positive {return (p1.x-p0.x) * (p2.y-p0.y)-(p2.x-p0.x) * (p1.y-p0.y);} void mo_hpi_addl (point a, point B) on the right) {Xian [Yong]. S = A; Xian [Yong]. E = B; Xian [Yong]. angle = atan2 (B. y-a.y, B. x-a.x); Yong ++;} // half plane bool mo_hpi_cmp (const Lin E & A, const line & B) {If (mo_ee (. angle, B. angle) {return mo_gg (mo_xmult (B. e,. s, B. s), 0);} else {return mo_ll (. angle, B. angle) ;}} int mo_hpi_dq [400 + 10]; bool mo_hpi_isout (line cur, line Top, line top_1) {point Jiao = mo_intersection (top. s, top. e, top_1.s, top_1.e); Return mo_ll (mo_xmult (cur. e, Jiao, cur. s), 0); // If the clockwise direction is mo_gg} int mo_halfplaneintersect (line * Xian, int N, point * Jiao) {int I, j, ret = 0; so RT (Xian, Xian + N, mo_hpi_cmp); for (I = 0, j = 0; I <n; I ++) {If (mo_gg (xian [I]. angle, Xian [J]. angle) {Xian [++ J] = Xian [I] ;}} n = J + 1; mo_hpi_dq [0] = 0; mo_hpi_dq [1] = 1; int Top = 1, bot = 0; for (I = 2; I <n; I ++) {While (top> bot & mo_hpi_isout (xian [I], xian [mo_hpi_dq [Top], Xian [mo_hpi_dq [Top-1]) Top --; while (top> bot & mo_hpi_isout (xian [I], xian [mo_hpi_dq [bot], Xian [mo_hpi_dq [bot + 1]) BOT ++; Mo_hpi_dq [++ top] = I; // current semi-plane inbound stack} while (top> bot & mo_hpi_isout (xian [mo_hpi_dq [bot], xian [mo_hpi_dq [Top], Xian [mo_hpi_dq [Top-1]) Top --; while (top> bot & mo_hpi_isout (xian [mo_hpi_dq [Top], xian [mo_hpi_dq [bot], Xian [mo_hpi_dq [bot + 1]) BOT ++; mo_hpi_dq [++ top] = mo_hpi_dq [bot]; for (ret = 0, I = Bot; I <top; I ++, RET ++) {Jiao [RET] = mo_intersection (xian [mo_hpi_dq [I + 1]. s, Xian [mo_hpi_dq [I + 1]. e, Xian [mo_hpi_dq [I]. s, Xian [mo_hpi_dq [I]. e);} return ret;} inline double mo_distance (point P1, point P2) {return SQRT (p1.x-p2.x) * (p1.x-p2.x) + (p1.y-p2.y) * (p1.y-p2.y);} // returns the point mo_getfaxian (point xiang) {point A; If (mo_ee (Xiang. x, 0) {. X = 1;. y = 0; return a;} else if (mo_ee (Xiang. y, 0) {. X = 0;. y = 1; return a;} else {. X = 1;. y =-1.0 * Xiang. x/Xiang. y; return a ;}/// the line se is pushed to a certain distance from line Mo _ Line_tui (point S, point E, double tui) {Line ret; point fa = mo_getfaxian (point (e. x-s.x, E. y-s.y); If (mo_ll (mo_xmult (E, point (s. X + FA. x, S. Y + FA. y), S), 0) {fa. X =-fa. x; fa. y =-fa. y;} double Len = SQRT (Fa. x * Fa. X + FA. y * Fa. y); Len = Len/Tui; fa. X = fa. x/Len; fa. y = fa. y/Len; S. X = S. X + FA. x, S. y = S. Y + FA. y; E. X = E. X + FA. x, E. y = E. Y + FA. y; ret. S = s; ret. E = E; return ret;} int main () {int I, j; while (scanf ("% d", & N )! = EOF) {Yong = 0; scanf ("% lf", & R); for (I = 0; I <n; ++ I) {scanf ("% lf", & Dian [I]. x, & Dian [I]. y) ;}for (I = 0; I <n; ++ I) {line LL = mo_line_tui (Dian [(I + 1) % N], Dian [I], r); mo_hpi_addl (LL. s, ll. e);} int ret = mo_halfplaneintersect (Xian, N, Jiao); double min =-1, min2; int A, B; for (I = 0; I <ret; ++ I) {for (j = I + 1; j <ret; ++ J) {min2 = mo_distance (Jiao [I], Jiao [J]); if (Min <0 | min2> min) {min = min2; A = I; B = J ;}} printf ("%. 4lf %. 4lf %. 4lf %. 4lf \ n ", Jiao [A]. x, Jiao [A]. y, Jiao [B]. x, Jiao [B]. y);} return 0 ;}