Poj 3301 Texas trip (three points)

Poj 3301 Texas trip (three points)

ACM

Question address:
Poj 3301 Texas trip

Question:
Given N points in a two-dimensional plane, a square with the smallest area is required to cover all points.

Analysis:
If you want to solve the problem, you will lose...
We can make the square do not move, and all vertices are rotated and transformed, so that the result will not be deformed.
Deformation means:x1=x*cos(a)-y*sin(a); y1=x*sin(a)+y*cos(a) 
I wanted to see it in a violent way. Later I found that it was three points, which proved to be a huge reference:

It can be proved that they are all convex functions, so the maximum difference between them is also a convex function. Therefore, we can use the three-way method to divide the Rotation Angle and find the smallest square that meets the requirements. For detailed procedures, see the code.

After sending a few rounds, I verified again that the double output of G ++ uses % F by default.

Code:

/**  Author:      illuz <iilluzen[at]gmail.com>*  File:        3301.cpp*  Create Date: 2014-09-18 09:25:04*  Descripton:  triple*/#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 31;const double PI = acos(-1.0);const double EPS = 1e-12;int t, n;double x[N], y[N];inline double calc(double a) {double minx = 1000, miny = 1000, maxx = -1000, maxy = -1000;double xx, yy;repf (i, 1, n) {xx = x[i] * cos(a) - y[i] * sin(a);yy = x[i] * sin(a) + y[i] * cos(a);minx = min(minx, xx);maxx = max(maxx, xx);miny = min(miny, yy);maxy = max(maxy, yy);}return max(maxx - minx, maxy - miny);}int main() {ios_base::sync_with_stdio(0);double l, r, mid, mmid, ans;cin >> t;while (t--) {cin >> n;repf (i, 1, n) {cin >> x[i] >> y[i];}l = 0.0;r = PI;while (r - l > EPS) {mid = (l + r) / 2;mmid = (mid + r) / 2;ans = calc(mid);if (ans <= calc(mmid))r = mmid;elsel = mid;}printf("%.2f\n", ans * ans);// brute force//ans = 1000;//for (double i = 0.0; i <= PI; i += 0.00005)//ans = min(ans, calc(i));//printf("%.2lf\n", ans * ans);}return 0;}

Poj 3301 Texas trip (three points)