题意：

给出平面上n个点，找一条直线，使得所有点在直线的同侧，且到直线的距离之平均值尽量小。 

先求凸包

易知最优直线一定是凸包的某条边，然后利用点到直线距离公式进行计算。

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;

struct Point {
    int x, y;
    Point(int x=0, int y=0):x(x),y(y) {}
};

typedef Point Vector;

Vector operator + (const Vector& a, const Vector& b) {
    return Vector(a.x+b.x, a.y+b.y);
}
Vector operator - (const Vector& a, const Vector& b) {
    return Vector(a.x-b.x, a.y-b.y);
}
Vector operator * (const Vector& a, double p) {
    return Vector(a.x*p, a.y*p);
}
Vector operator / (const Vector& a, double p) {
    return Vector(a.x/p, a.y/p);
}
bool operator < (const Point& p1, const Point& p2){
    return p1.x<p2.x ||(p1.x==p2.x&&p1.y<p2.y);
}

bool operator == (const Point& p1, const Point& p2){
    return p1.x == p2.x && p1.y == p2.y;
}

int Cross(const Vector& a, const Vector& b) {
    return a.x*b.y - a.y*b.x;
}

vector<Point> ConvexHull(vector<Point> p) {
    sort(p.begin(), p.end());
    p.erase( unique(p.begin(), p.end()), p.end());

    int n = p.size();
    int m = 0;
    vector<Point> ch(n+1);
    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--;
    ch.resize(m);
    return ch;
}

// 过两点p1, p2的直线一般方程ax+by+c=0
// (x2-x1)(y-y1) = (y2-y1)(x-x1)
void getLineGeneralEquation(const Point& p1, const Point& p2, double& a, double& b, double &c) {
  a = p2.y-p1.y;
  b = p1.x-p2.x;
  c = -a*p1.x - b*p1.y;
}

int main()
{
    int t, n, i, j;
    scanf("%d", &t);
    for(int cas=1; cas<=t; ++cas)
    {
        scanf("%d", &n);
        int x, y;
        vector<Point> P;
        double sumx = 0, sumy = 0;
        for(i=0; i<n; ++i)
        {
            scanf("%d%d", &x, &y);
            sumx += x;
            sumy += y;
            P.push_back(Point(x,y));
        }
        P = ConvexHull(P);
        int m = P.size();
        double ans = 1e9;
        if(m<=2) ans = 0;
        else
        for(i=0; i<m; ++i)
        {
                j = (i+1)%m;
                double A, B, C;
                getLineGeneralEquation(P[i], P[j], A, B, C);
                double tmp = fabs(A*sumx + B*sumy + C*n) / sqrt(A*A+B*B);
                ans = min(ans, tmp);
        }
        printf("Case #%d: %.3f\n", cas, ans/n);
    }
    return 0;
}