UVA 11168 Airport

There is a small town with n houses. The town needs an airport. An airport is basically a very long, very straight road. Think of it as an infinite line. We need to build the airport such that the average distance from each house to the airport is as small as possible. However, no one wants to walk across the runway, so all of the houses must be on the same side of the airport. (Some houses may be a distance of zero away from the runway, but that's ok; we'll give them some free ear plugs.)

Where should we build the airport, and what will be the average distance?

Input

The first line of input gives the number of cases, N (<=65) . N test cases follow. Each one is a line containing n (0 < n <= 10000) , followed by n lines giving the xy-coordinates of the houses. All coordinates are integers with absolute value of at most 80,000 .

Output

For each test case, output one line containing "Case #x:" followed by the average distance from the airport to the houses, with 3 digits after the decimal point. No answer will be within 10^-5 of a round-off error case.

Sample Input                             Output for Sample Input

4 4 0 0 0 1 1 0 1 1 2 15035 39572 34582 39535 3 0 0 0 1 1 0 5 0 0 0 2 2 0 2 2 11

Case #1: 0.500 Case #2: 0.000 Case #3: 0.236 Case #4: 1.000

给出平面上的n个点，求一条直线，使得所有点在该直线的同一侧且所有点到该直线的距离和最小，输出该距离和。

要使所有点在该直线的同一侧，明显是直接利用凸包的边更优。所以枚举凸包的每条边，然后求距离和。直线一般式为Ax + By + C = 0.点(x0, y0)到直线的距离为

fabs(Ax0+By0+C)/sqrt(A*A+B*B).由于所有点在直线的同一侧，那么对于所有点，他们的(Ax0+By0+C)符号相同，显然可以累加出sumX和sumY，然后统一求和。

#include <iostream>
#include <algorithm>
#include <math.h>
#include <iomanip>
using namespace std;
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y){}
} ;
bool operator < (const Point&a,const Point&b)
{
    return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
Point operator+(Point A,Point B)
{
    return Point(A.x+B.x,A.y+B.y);
}
Point operator-(Point A,Point B)
{
    return Point(A.x-B.x,A.y-B.y);
}
double Cross(Point A,Point B)
{
    return A.x*B.y-A.y*B.x;
}
void GetLineABC(Point A,Point B,double &a,double &b,double &c)//直线两点式转一般式
{
    a=A.y-B.y; b=B.x-A.x; c=A.x*B.y-A.y*B.x;
}
int ConvexHull(Point *p,int n,Point *ch)
{
    sort(p,p+n);
    int m=0;
    for(int i=0;i<n;i++)
    {
        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
        ch[m++]=p[i];
    }
    int k=m;
    for(int i=n-2;i>=0;i--)
    {
        while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
        ch[m++]=p[i];
    }
    if(n>1)m--;
    return m;
}


int main()
{
    int T;
    cin>>T;
    for(int cases=1;cases<=T;cases++)
    {
        int N;
        Point P[10010],ch[10010];
        cin>>N;
        double X=0,Y=0,ans=1e10;
        for(int i=0;i<N;i++)
        {
              cin>>P[i].x>>P[i].y;
              X=X+P[i].x;
              Y=Y+P[i].y;
        }
        int m=ConvexHull(P,N,ch);
        ch[m]=ch[0];
        for(int i=0;i<m;i++)
        {
            double a=0,b=0,c=0;
            GetLineABC(ch[i],ch[i+1],a,b,c);
            ans=min(ans,fabs(a*X+b*Y+c*N)/(sqrt(a*a+b*b)));
        }
        cout<<setiosflags(ios::fixed)<<setprecision(3);
        cout<<"Case #"<<cases<<": "<<(N>2?ans/N:0)<<endl;
    }
    return 0;
}