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
2
4
-1 -1
1 -1
1 1
-1 1
4
10 1
10 -1
-10 1
-10 -1

Sample Output
4.00
242.00
 

列舉旋轉角，逐步減小角度（每次讓其正切值減半）從而逼近最優解。
對於每一個旋轉角，設方向向量為(x, y)，那麼就可以借用點乘的幾何意義將原點到每個點的有向線段全部投影到(x, y)上，再用類似的方法將與(x, y)垂直的向量(y, -x)也計算一遍，兩者取較大，就可以將結果求出來。

標準的旋轉變換：

程式中使用的旋轉變換：

並且其中將向量的模長設為了100以減小相對誤差。
Accode：

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
const int maxN = 40;
const double INF = 1e198, zero = 1e-12;
const int dx[] = {1, -1}, dy[] = {-1, 1};
int x[maxN], y[maxN], n, T;

double calc(double _x, double _y)
{
    double Min = INF, Max = -INF;
    for (int i = 0; i < n; ++i)
    {
        double tmp = _x * x[i] + _y * y[i];
        if (tmp > Max) Max = tmp;
        if (tmp < Min) Min = tmp;
    }
    double ans = Max - Min;
    Min = INF, Max = -INF;
    for (int i = 0; i < n; ++i)
    {
        double tmp = _y * x[i] - _x * y[i];
        if (tmp > Max) Max = tmp;
        if (tmp < Min) Min = tmp;
    }
    (Max - Min > ans) ? (ans = Max - Min) : ans;
    return ans * ans / (_x * _x + _y * _y);
//記得要除以模長，防止開方，直接返回求得正方形的面積。
}

double SAA()
{
    double sx = 100, sy = 0, T = 1, ans = calc(sx, sy);
    for (; T > zero; T /= 2)
    while (1)
    {
        bool ok = 0;
        for (int i = 0; i < 2; ++i)
        {
            double tx = sx + dx[i] * sy * T,
                   ty = sy + dy[i] * sx * T,
                   Now = calc(tx, ty),
                   tmp = sqrt(1e4 / (tx * tx + ty * ty));
            tx *= tmp, ty *= tmp;
	//保持模長不變。
	    if (Now < ans)
            {ans = Now; ok = 1; sx = tx, sy = ty;}
	//這裡可以只接受更優的解，因為當前決策
	//並不影響下一步是否能達到最優解。
        } //向順時針和逆時針兩個方向都要旋轉一遍。
        if (!ok) break;
    }
    return ans;
}

int main()
{
    freopen("Texas_Trip.in", "r", stdin);
    freopen("Texas_Trip.out", "w", stdout);
    for (scanf("%d", &T); T; --T)
    {
        scanf("%d", &n);
        for (int i = 0; i < n; ++i)
            scanf("%d%d", x + i, y + i);
        printf("%.2lf\n", SAA());
    }
    return 0;
}