Poj3301--texas Trip (minimum square overlay)

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The main topic: given the coordinates of n points, now requires a square, completely surrounded by n points, and the smallest square area, the smallest square area.

The expression cannot understand why the area changes with the angle is a single-peak function, waiting for Daniel to tell you that,

If the area changes with the angle is a single-peak function, then naturally you can think of three points, according to the question asked to find the smallest square area, if the square is parallel to the x-axis, then the square area is the maximum distance of x *y maximum distance. Then rotate the square, in 0-90 degrees will always find a square area of the minimum,, but the rotation of the square is more troublesome, we can consider rotating coordinate system, will do coordinate system rotation 0-90 degrees, according to the angle of rotation to recalculate the coordinates of the points, and then find the difference between the X and Y, calculate the area.

Because it is a single-peak function, it uses a three-point angle to find a minimum area.

Note: The eqs of the three points to be very small,,,,,

Angular rotation Formula x = X*cos (j)-Y*sin (j); y = X*sin (j) + Y*cos (j);

#include <cstdio> #include <cstring> #include <cmath> #include <algorithm>using namespace std; Define EQS 1e-12#define INF 0x3f3f3f3f#define PI acos ( -1.0) struct node{double x, y;}    P[35];int N;d ouble Maxx, Minx, Maxy, Miny;d ouble F (double j) {int i;    Double x, y;    Maxx = Maxy =-inf;    Minx = Miny = INF;        for (i = 0; i < n; i++) {x = P[i].x*cos (j)-P[i].y*sin (j);        y = P[i].x*sin (j) + P[i].y*cos (j);        Maxx = max (maxx,x);        Minx = min (minx,x);        Maxy = max (maxy,y);    miny = min (miny,y); } return Max (Maxx-minx,maxy-miny);}    int main () {int T, I;    scanf ("%d", &t);        while (t--) {scanf ("%d", &n);        for (i = 0; i < n; i++) {scanf ("%lf%lf", &p[i].x, &AMP;P[I].Y);        } Double low = 0.0, MID1, mid2, high = pi/2.0;            while (Low + EQs < high) {MID1 = (low + high)/2.0;   Mid2 = (mid1 + high)/2.0;         if (f (MID1) > F (mid2)) low = MID1;        else high = Mid2;        } low = f (low);    printf ("%.2lf\n", Low*low); } return 0;}

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Poj3301--texas Trip (minimum square overlay)