poj3301 三分

Texas Trip

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4998		Accepted: 1559

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
题目大意：给你平面上的若干个点的坐标，让你求能将这些点覆盖的正方形的最小面积是多少。
思路分析：刚开始我一直搞不懂第二组样例怎么来的，我一直觉得是400，后来才想明白正方形是可以进行旋转的，
然后解题思路又好久没有想明白，这道题在题目分类是三分，一直不清楚到底应该怎么进行三分，后来想了一下
正方形的旋转不容易解决，但是我可以让坐标轴旋转啊，很明显在坐标轴旋转的过程中最小正方形的边长是先减后
增的趋势，为单峰函数，因此可以用三分法来进行解决。
代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int maxn=35;
#define eps 1e-8
const int inf=0xfffffff;
const double pi=acos(-1.0);
struct nod
{
    int x;
    int y;
};
nod point[maxn];
int n;
double cul(double a)
{
    double minx=inf,maxx=-inf;
    double miny=inf,maxy=-inf;
    for(int i=0;i<n;i++)
    {
        double xx=point[i].x*cos(a)+point[i].y*sin(a);
        double yy=point[i].y*cos(a)-point[i].x*sin(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()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d%d",&point[i].x,&point[i].y);
        double l=0.0,r=pi;
        double ans;
        while(l+eps<=r)
        {
            double mid=(l+r)/2;
            double mmid=(mid+r)/2;
            double t1=cul(mid),t2=cul(mmid);
            if(t1<=t2) r=mmid;
            else l=mid;
        }
        printf("%.2lf\n",cul(l)*cul(l));
    }
    return 0;
}