Question:
There are N points on the plane. Find a straight line so that all points are on the same side of the straight line. And find the minimum value of the distance between these points and the straight line.
Analysis:
As long as the straight line does not pass through the convex hull, the first condition is met. To minimize the distance and distance, the straight line must be on the side of the convex hull. After finding the convex hull, enumerate each side and find the sum of the distance from all points to the straight line to get the minimum value.
The formula for distance from a point to a straight line is:
Because the points are on the same side of a straight line, we can "move" the addition to it, and finally obtain the absolute value. Therefore, we can preprocess the sum of the X and Y coordinates of all points. Of course, the constant C must also be multiplied by N times.
We know that the two-point coordinate is used to find the equation of the straight line at this point. This is a good solution. Considering that the straight line does not have a slope, We need to multiply the denominator in the expression.
1 // # define local 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <cmath> 6 # include <vector> 7 using namespace std; 8 9 struct point 10 {11 double X, Y; 12 point (double x = 0, Double Y = 0): x (x), y (y) {} 13 }; 14 typedef point vector; 15 point operator + (point a, point B) 16 {17 return point (. X + B. x,. Y + B. y); 18} 19 point operator-(point a, point B) 20 {21 return point (. x-B.x,. y-B.y); 22} 23 bool operator <(const point & A, const point & B) 24 {25 return. x <B. X | (. X = B. X &. Y <B. y); 26} 27 bool operator = (const point & A, const point & B) 28 {29 return. X = B. X &. y = B. y; 30} 31 double cross (vector A, vector B) 32 {33 return. x * B. y-. y * B. x; 34} 35 36 vector <point> convexhull (vector <point> P) {37 // preprocessing, delete repeat 38 sort (P. begin (), P. end (); 39 p. erase (unique (P. begin (), P. end (), p. end (); 40 41 int n = P. size (); 42 int m = 0; 43 vector <point> CH (n + 1); 44 for (INT I = 0; I <n; I ++) {45 while (M> 1 & cross (CH M-1]-ch [m-2], p [I]-ch [m-2]) <= 0) m --; 46 ch [M ++] = P [I]; 47} 48 int K = m; 49 for (INT I = n-2; I> = 0; I --) {50 while (M> K & cross (CH M-1]-ch [m-2], p [I]-ch [m-2]) <= 0) m --; 51 ch [M ++] = P [I]; 52} 53 If (n> 1) m --; 54 // For (INT I = 0; I <m; ++ I) printf ("% lf \ n", CH [I]. x, CH [I]. y); 55 ch. resize (m); 56 return ch; 57} 58 59 double sumx, Sumy; 60 61 double dist (point a, point B, int m) 62 {63 double A =. y-b.y, B = B. x-a.x, c =. x * B. y-B. x *. y; 64 // printf ("% lf", FABS (A * sumx + B * Sumy + C), SQRT (A * A + B * B )); 65 return (FABS (A * sumx + B * Sumy + C * m)/SQRT (A * A + B * B); 66} 67 68 int main (void) 69 {70 # ifdef local 71 freopen ("11168in.txt", "r", stdin); 72 # endif 73 74 int t; 75 scanf ("% d", & T ); 76 for (INT Kase = 1; Kase <= T; ++ Kase) 77 {78 int N; 79 vector <point> P; 80 sumx = 0.0, Sumy = 0.0; 81 scanf ("% d", & N); 82 For (INT I = 0; I <n; ++ I) 83 {84 Double X, Y; 85 scanf ("% lf", & X, & Y); 86 p. push_back (point (x, y); 87 sumx + = x; Sumy + = y; 88} 89 vector <point> CH = convexhull (P ); 90 int M = CH. size (); 91 // For (INT I = 0; I <m; ++ I) printf ("% lf \ n", CH [I]. x, CH [I]. y); 92 If (M <= 2) 93 {94 printf ("case # % d: 0.000 \ n", Kase); 95 continue; 96} 97 98 double ans = 1e10; 99 for (INT I = 0; I <m ;++ I) 100 Ans = min (ANS, dist (CH [I], ch [(I + 1) % m], n); 101 printf ("case # % d: %. 3lf \ n ", Kase, ANS/n); 102} 103}
Code Jun
Ultraviolet A 11168 (convex bag + distance from a point to a straight line) Airport