Ultraviolet A 11168

Question: There are N points. Find a straight line so that all points are on the same side of a straight line (or in a straight line), and the distance to the straight line is the smallest. The ratio of the output distance to the number of points.
Question: If all vertices are located on the same side of a straight line, that is, the straight line cannot pass through the convex hull, the straight line where the convex hull edge is located is a possible solution. Point (x0, y0) distance to a straight line (AX + by + c = 0)
Dis = | ax0 + by0 + c |/SQRT (a ^ 2 + B ^ 2)
Calculate the distance and distance between all vertices. Based on the formula, calculate the sum of the X and Y coordinates of all vertices to reduce the time complexity.

# Include <cstdio> # include <cstring> # include <cmath> # include <algorithm> using namespace STD; const double Pi = ACOs (-1); const int n = 10005; const double INF = 1e9; struct point {Double X, Y; point (double x = 0, Double Y = 0): x (x), y (y) {} p [N], Res [N]; struct circle {point C; Double R; Circle () {} circle (point c, double r = 0 ): C (c), R (r) {} Point (double A) {return point (c. X + cos (a) * R, C. Y + Sin (a) * C. y) ;}}; int N; double sqr (Double X) {return x * X;} point operator + (point a, point B) {return point (. X + B. x,. Y + B. y);} point operator-(point a, point B) {return point (. x-B. x,. y-B. y);} point operator * (point a, Double P) {return point (. x * P,. y * P);} point operator/(point a, Double P) {return point (. x/P,. y/P);} // calculate the positive and negative values of the dot product. The angle between the two values is an empty angle int DCMP (Double X) {If (FABS ). (X) <1e-9) return 0; return x <0? -1: 1;} bool operator <(const point & A, const point & B) {return. x <B. X | (. X = B. X &. Y <B. y);} bool operator = (const point & A, const point & B) {return DCMP (. x-B. x) = 0 & DCMP (. y-B. y) = 0;} // calculate the dot product double dot (point a, point B) {return. x * B. X +. y * B. y;} // calculate the cross product, that is, the number product double cross (point a, point B) {return. x * B. y-. y * B. X ;}// returns the vector length double length (point a) {return SQRT (dot (A, A);} // vector A rotates the rad radian, the negative value of rad is clockwise rotation point rotate (point a, double rad) {return point (. x * Cos (RAD)-. y * sin (RAD),. x * sin (RAD) +. y * Cos (RAD);} // converts degrees to radians double torad (double deg) {return deg/180.0 * PI;} int convexhull (point * P, Int & CNT, point * res) {sort (p, p + CNT); CNT = unique (p, p + CNT)-P; int m = 0; For (INT I = 0; I <CNT; I ++) {While (M> 1 & cross (RES [m-1]-res [m-2], P [I]-res [m-2]) <= 0) m --; Res [M ++] = P [I];} int K = m; for (INT I = CNT-2; I> = 0; I --) {While (M> K & cross (RES [m-1]-res [m-2], p [I]-res [m-2]) <= 0) M --; Res [M ++] = P [I];} If (CNT> 1) m --; return m;} void change (double & A, double & B, double & C, point A, point B) {A = B. y-. y; B =. x-B. x; C = B. x *. y-. x * B. y;} int main () {int T, CAS = 1; scanf ("% d", & T); While (t --) {scanf ("% d ", & N); double sumx = 0, Sumy = 0; For (INT I = 0; I <n; I ++) {scanf ("% lf ", & P [I]. x, & P [I]. y); sumx + = P [I]. x; Sumy + = P [I]. y;} If (n <= 2) {printf ("case # % d: 0.000 \ n", CAS ++); continue;} int temp = convexhull (p, n, Res); double Minn = inf; For (INT I = 0; I <temp; I ++) {double A, B, C; Change (A, B, c, res [I], Res [(I + 1) % TEMP]); double sumdis = FABS (A * sumx + B * Sumy + C * n) /SQRT (sqr (A) + sqr (B); Minn = min (Minn, sumdis);} printf ("case # % d: %. 3lf \ n ", CAS ++, Minn/n);} return 0 ;}

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Ultraviolet A 11168