POJ 3301 Texas Trip(三分)

bigbigship發表於2014-07-29

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
給定n個點的座標求一個最小的正方形使n個點都在正方形內
先假定邊長最小且個邊都平行於座標軸的正方形，然後三分旋轉的角度 
這裡要知道座標旋轉公式：
如果為旋轉前的座標，為旋轉後的座標，那麼有：
 
程式碼如下：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double eps = 1e-12;

const double INF = 1e6;

const double PI = acos(-1.0);

double p[35][2];

int n;

double cir(double x){//旋轉x角度後對應點的位置
    double minx,miny,maxx,maxy;
    double tmpx,tmpy;
    minx=miny=INF;
    maxx=maxy=-INF;
    for(int i=0;i<n;i++){
        tmpx=p[i][0]*cos(x)-p[i][1]*sin(x);
        tmpy=p[i][1]*cos(x)+p[i][0]*sin(x);
        minx=min(tmpx,minx);
        miny=min(tmpy,miny);
        maxx=max(tmpx,maxx);
        maxy=max(tmpy,maxy);
    }
    return max(maxx-minx,maxy-miny);
}

double ternary_search(double l,double r){//三分旋轉的角度
    double ll,rr;
    while(r-l>eps){
        ll=(l*2+r)/3;
        rr=(l+r*2)/3;
        if(cir(ll)>cir(rr))
            l=ll;
        else
            r=rr;
    }
    return l;
}

int main()
{
    int cas;
    cin>>cas;
    while(cas--){
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>p[i][0]>>p[i][1];
        double l=0,r=PI/2;
        double tmp=ternary_search(l,r);
        printf("%.2lf\n",cir(tmp)*cir(tmp));
    }
    return 0;
}




            




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