uva 11168

题意：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。

思路：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。所以枚举凸包的没条边，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。

水。。。。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<vector>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long int
#define up(i,x,y) for(i=x;i<=y;i++)
#define w(a) while(a)
using namespace std;
const double inf=0x3f3f3f3f;
const int N = 4010;
const double eps = 5*1e-13;
const double PI = acos(-1.0);

double torad(double deg)
{
    return deg/180 * PI;
}

struct Point
{
    double x, y;
    Point(double x=0, double 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 Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}
double Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}

Vector Rotate(const Vector& A, double rad)
{
    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}

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;
}

// 点集凸包
// 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
// 如果不介意点集被修改，可以改成传递引用
vector<Point> ConvexHull(vector<Point> p)
{
    // 预处理，删除重复点
    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;
    double x,y;
    double sumx,sumy;
    double ans;
    int cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        vector<Point>p;
        sumx=0;sumy=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&x,&y);
            sumx+=x;sumy+=y;
            p.push_back(Point(x,y));
        }
        vector<Point>ch=ConvexHull(p);
        ans=inf;
        int edge=ch.size();
        if(edge<=2)
        {
            ans=0;
        }
        else
        {
            for(int i=0;i<edge;i++)
            {
                double a,b,c;
                getLineGeneralEquation(ch[i],ch[(i+1)%edge],a,b,c);
                ans=min(ans,fabs(sumx*a+sumy*b+n*c)/sqrt(a*a+b*b));
            }
        }
        printf("Case #%d: %.3f
",cas++,ans/n);
    }
    return 0;
}