UVA 11168

题目链接：https://cn.vjudge.net/problem/UVA-11168 

题意：

给出平面上的n个点，求一条直线，使得所有的点在该直线的同一侧（可以在该直线上），并且所有点到该直线的距离和最小，输出该距离除以n；

题解：

显然最好能让越多的点在这条直线上就越好，但又要所有点满足在同侧，则显然要选取某条凸包边界所在直线作为ans；

求出凸包后，遍历每条凸包边，求出所有点到这条直线的距离和，找到最小的即可。

AC代码：

#include<bits/stdc++.h>
#define MAX 10005
using namespace std;

//--------------------------------------计算几何模板 - st--------------------------------------

const double eps = 1e-6;

struct Point{
    double x,y;
    Point(double tx=0,double ty=0):x(tx),y(ty){}
};
typedef Point Vctor;

//向量的加减乘除
Vctor operator + (Vctor A,Vctor B){return Vctor(A.x+B.x,A.y+B.y);}
Vctor operator - (Point A,Point B){return Vctor(A.x-B.x,A.y-B.y);}
Vctor operator * (Vctor A,double p){return Vctor(A.x*p,A.y*p);}
Vctor operator / (Vctor A,double p){return Vctor(A.x/p,A.y/p);}
bool operator < (Point A,Point B){return A.x < B.x || (A.x == B.x && A.y < B.y);}

struct Line{
    Point p;
    Vctor v;
    Line(Point p=Point(0,0),Vctor v=Vctor(0,0)):p(p),v(v){}
    Point point(double t){return p + v*t;} //获得直线上的距离p点t个单位长度的点
};
struct Circle{
    Point c;
    double r;
    Circle(Point tc=Point(0,0),double tr=0):c(tc),r(tr){}
    Point point(double a){return Point( c.x + cos(a)*r , c.y + sin(a)*r);}
};

int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return (x<0)?(-1):(1);
}
bool operator == (Point A,Point B){return dcmp(A.x-B.x)==0 && dcmp(A.y-B.y)==0;}

//向量的点积，长度，夹角
double Dot(Vctor A,Vctor B){return A.x*B.x+A.y*B.y;}
double Length(Vctor A){return sqrt(Dot(A,A));}
double Angle(Vctor A,Vctor B){return acos(Dot(A,B)/Length(A)/Length(B));}

//叉积，三角形面积
double Cross(Vctor A,Vctor B){return A.x*B.y-A.y*B.x;}
double TriangleArea(Point A,Point B,Point C){return Cross(B-A,C-A);}

//向量的旋转,求向量的单位法线（即左转90度，然后长度归一）
Vctor Rotate(Vctor A,double rad){return Vctor( A.x*cos(rad) - A.y*sin(rad) , A.x*sin(rad) + A.y*cos(rad) );}
Vctor Normal(Vctor A)
{
    double L = Length(A);
    return Vctor(-A.y/L, A.x/L);
}

//直线的交点
Point getLineIntersection(Line L1,Line L2)
{
    Vctor u = L1.p-L2.p;
    double t = Cross(L2.v,u)/Cross(L1.v,L2.v);
    return L1.p + L1.v*t;
}

//点到直线的距离
double DistanceToLine(Point P,Line L)
{
    return fabs(Cross(P-L.p,L.v))/Length(L.v);
}

//点到线段的距离
double DistanceToSegment(Point P,Point A,Point B)
{
    if(A==B) return Length(P-A);
    Vctor v1 = B-A, v2 = P-A, v3 = P-B;
    if (dcmp(Dot(v1,v2)) < 0) return Length(v2);
    else if (dcmp(Dot(v1,v3)) > 0) return Length(v3);
    else return fabs(Cross(v1,v2))/Length(v1);
}

//点到直线的映射
Point getLineProjection(Point P,Line L)
{
    return L.v + L.v*Dot(L.v,P-L.p)/Dot(L.v,L.v);
}

//判断线段是否规范相交
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
    double c1 = Cross(a2 - a1,b1 - a1), c2 = Cross(a2 - a1,b2 - a1),
           c3 = Cross(b2 - b1,a1 - b1), c4 = Cross(b2 - b1,a2 - b1);
    return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}

//判断点是否在一条线段上
bool OnSegment(Point P,Point a1,Point a2)
{
    return dcmp(Cross(a1 - P,a2 - P))==0 && dcmp(Dot(a1 - P,a2 - P))<0;
}

//多边形面积
double PolgonArea(Point *p,int n)
{
    double area=0;
    for(int i=1;i<n-1;i++) area += Cross( p[i]-p[0] , p[i + 1]-p[0] );
    return area/2;
}

//判断圆与直线是否相交以及求出交点
int getLineCircleIntersection(Line L,Circle C,vector<Point> &sol)
{
    double t1,t2;
    double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
    double e = a*a + c*c , f = 2*(a*b + c*d),  g = b*b + d*d - C.r*C.r;
    double delta = f*f - 4.0*e*g;
    if(dcmp(delta)<0) return 0;
    else if(dcmp(delta)==0)
    {
        t1 = t2 = -f/(2.0*e);
        sol.push_back(L.point(t1));
        return 1;
    }
    else
    {
        t1 = (-f-sqrt(delta))/(2.0*e); sol.push_back(L.point(t1));
        t2 = (-f+sqrt(delta))/(2.0*e); sol.push_back(L.point(t2));
        return 2;
    }
}

//判断并求出两圆的交点
double angle(Vctor v){return atan2(v.y,v.x);}
int getCircleIntersection(Circle C1,Circle C2,vector<Point> &sol)
{
    double d = Length(C1.c - C2.c);
    //圆心重合
    if(dcmp(d)==0)
    {
        if(dcmp(C1.r-C2.r)==0) return -1; //两圆重合
        else return 0; //包含关系
    }

    //圆心不重合
    if(dcmp(C1.r+C2.r-d)<0) return 0; // 相离
    if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0; // 包含

    double a = angle(C2.c - C1.c);
    double da = acos((C1.r*C1.r + d*d - C2.r*C2.r) / (2*C1.r*d));
    Point p1 = C1.point(a - da), p2 = C1.point(a + da);
    sol.push_back(p1);
    if(p1==p2) return 1;
    sol.push_back(p2);
    return 2;
}

//求点到圆的切线
int getTangents(Point p,Circle C,vector<Line> &sol)
{
    Vctor u=C.c-p;
    double dis=Length(u);
    if(dis<C.r) return 0;
    else if(dcmp(dis-C.r) == 0)
    {
        sol.push_back(Line(p,Rotate(u,M_PI/2)));
        return 1;
    }
    else
    {
        double ang=asin(C.r/dis);
        sol.push_back(Line(p,Rotate(u,-ang)));
        sol.push_back(Line(p,Rotate(u,ang)));
        return 2;
    }
}

//求两圆的切线
int getCircleTangents(Circle A,Circle B,Point *a,Point *b)
{
    int cnt = 0;
    if(A.r<B.r){swap(A,B);swap(a,b);}
    //圆心距的平方
    double d2 = (A.c.x - B.c.x)*(A.c.x - B.c.x) + (A.c.y - B.c.y)*(A.c.y - B.c.y);
    double rdiff = A.r - B.r;
    double rsum = A.r + B.r;
    double base = angle(B.c - A.c);
    //重合有无限多条
    if(d2 == 0 && dcmp(A.r - B.r) == 0) return -1;
    //内切
    if(dcmp(d2-rdiff*rdiff) == 0)
    {
        a[cnt] = A.point(base);
        b[cnt] = B.point(base);
        cnt++;
        return 1;
    }
    //有外公切线
    double ang = acos((A.r - B.r) / sqrt(d2));
    a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
    a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;

    //一条内切线
    if(dcmp(d2-rsum*rsum) == 0)
    {
        a[cnt] = A.point(base);
        b[cnt] = B.point(M_PI + base);
        cnt++;
    }//两条内切线
    else if(dcmp(d2-rsum*rsum) > 0)
    {
        double ang = acos((A.r + B.r) / sqrt(d2));
        a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++;
        a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++;
    }
    return cnt;
}

//--------------------------------------计算几何模板 - ed--------------------------------------
Point p[MAX];

bool cmp(Point p1,Point p2)
{
    double tmp=Cross(p1-p[0],p2-p[0]);
    if(!dcmp(tmp)) return Length(p1-p[0])<Length(p2-p[0]);
    else return tmp>0;
}
vector<Point> graham_scan(int n)
{
    vector<Point> ans;
    int idx=0;
    for(int i=1;i<n;i++)//选出Y坐标最小的点，若Y坐标相等，选择X坐标小的点
    {
        if(p[i].y<p[idx].y || (p[i].y == p[idx].y && p[i].x < p[idx].x)) idx=i;
    }
    swap(p[0],p[idx]);
    sort(p+1,p+n,cmp);
    for(int i=0;i<=2;i++) ans.push_back(p[i]);
    int top=2;
    for(int i=3;i<n;i++)
    {
        while(top>0 && Cross(p[i]-ans[top-1],ans[top]-ans[top-1]) >= 0)
        {
            ans.pop_back();
            top--;
        }
        ans.push_back(p[i]);
        top++;
    }
    return ans;
}

int n;
int main()
{
    int t;
    scanf("%d",&t);
    for(int kase=1;kase<=t;kase++)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
        vector<Point> ans=graham_scan(n);
        double mini=0x3f3f3f3f;
        for(int i=0;i<ans.size();i++)
        {
            Point& p1 = ans[i];
            Point& p2 = (i==ans.size()-1)?ans[0]:ans[i+1];
            double dist=0;
            for(int i=0;i<n;i++) dist+=DistanceToLine(p[i],Line(p1,p2-p1));
            if(dist<mini) mini=dist;
        }
        printf("Case #%d: %.3lf
",kase,mini/(1.0*n));
    }
}

PS.为保证跟前面的计算几何模板一致性，就把整个模板都copy了。