POJ 3301 Texas Trip

神奇的3分法，求单峰函数极值的利器！！！
公式可以画图推导如下：

x1=cos(a)*x+sin(a)*y;

y1=cos(a)*y-sin(a)*x;

Texas Trip

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3429		Accepted: 1018

Description

After a day trip with his friend Dick, Harry noticed a strange pattern of tiny holes in the door of his SUV. The local American Tire store sells fiberglass patching material only in square sheets. What is the smallest patch that Harry needs to fix his door?

Assume that the holes are points on the integer lattice in the plane. Your job is to find the area of the smallest square that will cover all the holes.

Input

The first line of input contains a single integer T expressed in decimal with no leading zeroes, denoting the number of test cases to follow. The subsequent lines of input describe the test cases.

Each test case begins with a single line, containing a single integer n expressed in decimal with no leading zeroes, the number of points to follow; each of the following n lines contains two integers x and y, both expressed in decimal with no leading zeroes, giving the coordinates of one of your points.

You are guaranteed that T ≤ 30 and that no data set contains more than 30 points. All points in each data set will be no more than 500 units away from (0,0).

Output

Print, on a single line with two decimal places of precision, the area of the smallest square containing all of your points.

Sample Input

24-1 -11 -11 1-1 1410 110 -1-10 1-10 -1

Sample Output

4.00242.00

Source

WaterlooLocalContest, 2007.7.14

用C++提交可以AC，但用G++交WA，可能是精度处理不同～～～

  1 #include <cstdio>
  2 #include <cmath>
  3 
  4 using namespace std;
  5 
  6 #define INF (1<<25)
  7 #define eps (1e-12)
  8 #define PI acos(-1.0);
  9 
 10 double x[33];
 11 double y[33];
 12 
 13 int n;
 14 
 15 double mindis(double a)
 16 {
 17     double yMin=INF*1.0,xMin=INF*1.0,xMax=-INF*1.,yMax=-INF*1.0;
 18     double xx[33];
 19     double yy[33];
 20 
 21     for(int i=0;i<n;i++)
 22     {
 23         xx[i]=x[i];
 24         yy[i]=y[i];
 25     }
 26 
 27     for(int i=0;i<n;i++)
 28     {
 29         double xk=xx[i],yk=yy[i];
 30         xx[i]=cos(a)*xk+sin(a)*yk;
 31         yy[i]=cos(a)*yk-sin(a)*xk;
 32     }
 33 
 34     for(int i=0;i<n;i++)
 35     {
 36         if(xx[i]<xMin)
 37         {
 38             xMin=xx[i];
 39         }
 40         if(yy[i]<yMin)
 41         {
 42             yMin=yy[i];
 43         }
 44         if(xx[i]>xMax)
 45         {
 46             xMax=xx[i];
 47         }
 48         if(yy[i]>yMax)
 49         {
 50             yMax=yy[i];
 51         }
 52     }
 53 
 54   double  ansx=xMax-xMin;
 55   double  ansy=yMax-yMin;
 56 
 57     if(ansx-ansy>eps)
 58     {
 59         return ansx;
 60     }
 61     else
 62         return ansy;
 63 
 64 }
 65 
 66 
 67 int main()
 68 {
 69     int m;
 70     scanf("%d",&m);
 71     while(m--)
 72     {
 73         scanf("%d",&n);
 74         for(int i=0;i<n;i++)
 75         {
 76             scanf("%lf%lf",&x[i],&y[i]);
 77         }
 78 
 79         double st=0,ed=PI;
 80 
 81         while(ed-st>=eps)
 82         {
 83             double m1=(st*2+ed)/3.;
 84             double m2=(st+2*ed)/3.;
 85 
 86             double dism1=mindis(m1);
 87             double dism2=mindis(m2);
 88 
 89             if(dism1-dism2<eps)
 90             {
 91                 ed=m2;
 92             }
 93             else
 94             {
 95                 st=m1;
 96             }
 97 
 98         }
 99 
100         printf("%.2lf\n",mindis(st)*mindis(st));
101     }
102 
103     return 0;
104 }