POJ 3301 三分（最小覆盖正方形）

      想一下，如果题目中规定正方形必须和x轴平行，那么我们是不是直接找到最大的x差和最大的y差取最大就行了，但是这个题目没说平行，那么我们就旋转这个正方形，因为是凸性（或者凹性）用三分去枚举正方形的角度[0,PI/2],然后缩小范围，知道找到答案。公式是

nowx = x * cos(du) - y * sin(d)    nowy = x * sin(du)  + y *cos(d)

#include<stdio.h>
#include<math.h>

#define N 50
#define eps 0.000001

double PI = acos(-1.0);

typedef struct
{
   double x ,y;
}NODE;

NODE node[N];

double maxx(double x ,double y)
{
    return x > y ? x : y;
}

double minn(double x ,double y)
{
    return x < y ? x : y;
}

double abss(double x)
{
   return x < 0 ? -x : x;
}

double now(double phi ,int n)
{
   double Max_x = -100000000 ,Min_x = 100000000;
   double Max_y = -100000000 ,Min_y = 100000000;
   for(int i = 1 ;i <= n ;i ++)
   {
      double xx = node[i].x * cos(phi) - node[i].y * sin(phi);
      double yy = node[i].x * sin(phi) + node[i].y * cos(phi);
      Max_x = maxx(xx ,Max_x);
      Max_y = maxx(yy ,Max_y);
      Min_x = minn(Min_x ,xx);
      Min_y = minn(Min_y ,yy);
   }
   return maxx((Max_x - Min_x) ,(Max_y - Min_y));
}

int main ()
{
   int t ,n ,i;
   double low ,mid ,mmid ,up ,ans;
   scanf("%d" ,&t);
   while(t--)
   {
      scanf("%d" ,&n);
      for(i = 1 ;i <= n ;i ++)
      scanf("%lf %lf" ,&node[i].x ,&node[i].y);
      low = 0 ,up = PI / 2.0;
      while(1)
      {
         mid = (low + up) / 2.0;
         mmid = (mid + up) / 2.0;
         double dis1 = now(mid ,n);
         double dis2 = now(mmid ,n);
         if(dis1 < dis2) up = mmid;
         else  low = mid;
         if(abss(dis1 - dis2) < eps)
         break;
         ans = minn(dis1 ,dis2);
      }
      printf("%.2lf\n" ,ans * ans);
   }
   return 0;
}