Poj 3130 How I Mathematician Wonder What You Are! -Check whether the polygon has cores.

/* Poj 3130 How I Mathematician Wonder What You Are! -Check whether the polygon has cores */# include <stdio. h> # include <math. h> const double eps = 1e-8; const int N = 103; struct point {double x, y;} dian [N]; inline bool mo_ee (double x, double y) {double ret = x-y; if (ret <0) ret =-ret; if (ret <eps) return 1; return 0;} inline bool mo_gg (double x, double y) {return x> y + eps;} // x> yinline bool mo_ll (double x, double y) {return x <y-eps ;} // x <yinline bool mo_ge (double x, double y) {return x> y-eps;} // x> = yinline bool mo_le (double x, double y) {return x <y + eps;} // x <= yinline double mo_xmult (point p2, point p0, point p1) // p1 returns a negative value on the left of p2, return the positive {return (p1.x-p0.x) * (p2.y-p0.y)-(p2.x-p0.x) * (p1.y-p0.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 ;} /////////////// // obtain the semi-plane intersection point mo_banjiao_jiao [N * 2] by the cutting method. point mo_banjiao_jiao_temp [N * 2]; void mo_banjiao_cut (point * ans, point qian, point hou, int & nofdian) {int I, k; for (I = k = 0; I <nofdian; ++ I) {double a, B; a = mo_xmult (hou, ans [I], qian); B = mo_xmult (hou, ans [(I + 1) % nofdian], qian); if (mo_ge (a, 0 )) // clockwise <= 0 {mo_banjiao_jiao_temp [k ++] = ans [I];} if (mo_ll (a * B, 0 )) {mo_banjiao_jiao_temp [k ++] = mo_intersection (qian, hou, ans [I], ans [(I + 1) % nofdian]) ;}} for (I = 0; I <k; ++ I) {ans [I] = mo_banjiao_jiao_temp [I];} nofdian = k;} int mo_banjiao (point * dian, int n) {int I, nofdian; nofdian = n; for (I = 0; I <n; ++ I) {mo_banjiao_jiao [I] = dian [I];} for (I = 0; I <n; ++ I) // I starts from 0 {mo_banjiao_cut (mo_banjiao_jiao, dian [I], dian [(I + 1) % n], nofdian ); if (nofdian = 0) {return nofdian ;} /// // int main () {int t, I, n; while (scanf ("% d", & n), n) {for (I = 0; I <n; ++ I) {scanf ("% lf ", & dian [I]. x, & dian [I]. y);} int ret = mo_banjiao (dian, n); if (ret = 0) {printf ("0 \ n ");} else {printf ("1 \ n") ;}} return 0 ;}

/* Why ret <3? */# Include <stdio. h> # include <math. h ># include <algorithm> using namespace std; const double eps = 1e-8; struct point {double x, y;} dian [20000 + 10]; point jiao [203]; struct line {point s, e; double angle;} xian [20000 + 10]; int n, yong; bool mo_ee (double x, double y) {double ret = x-y; if (ret <0) ret =-ret; if (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 line & 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 [20000 + 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; sort (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 half-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;} int main () {int I; while (scanf ("% d", & n), n) {yong = 0; for (I = 0; I <n; ++ I) {scanf ("% lf", & dian [I]. x, & dian [I]. y) ;}for (I = 0; I <n; ++ I) {mo_HPI_addl (dian [I], dian [(I + 1) % n]);} int ret = mo_HalfPlaneIntersect (xian, n, jiao); if (ret <3) {printf ("0 \ n ");} else {printf ("1 \ n") ;}} return 0 ;}