UVA 11168 Airport(凸包)

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。

#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<fstream>#include<sstream>#include<bitset>#include<vector>#include<string>#include<cstdio>#include<cmath>#include<stack>#include<queue>#include<stack>#include<map>#include<set>#define FF(i, a, b) for(int i=a; i<b; i++)#define FD(i, a, b) for(int i=a; i>=b; i--)#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define debug puts("**debug**")#define LL long long#define PB push_back#define eps 1e-10using namespace std;struct Point{    double x, y;    Point (double x=0, double y=0):x(x), y(y) {}};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }bool operator < (const Point& a, const Point& b){    return a.x < b.x || (a.x == b.x && a.y < b.y);}int dcmp(double x){    if(fabs(x) < eps) return 0;    return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b){    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }double Length(Vector A) { return sqrt(Dot(A, A)); }double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }double Area2(Vector A, Vector B, Vector C) { return Cross(B-A, C-A); }//向量逆时针旋转Vector Rotate(Vector A, double rad){    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));}//求直线p+tv和q+tw的交点 Cross(v, w) == 0无交点Point GetLineIntersection(Point p, Vector v, Point q, Vector w){    Vector u = p-q;    double t = Cross(w, u) / Cross(v, w);    return p + v*t;}//点p到直线ab的距离double DistanceToLine(Point p, Point a, Point b){    Vector v1 = b - a, v2 = p - a;    return fabs(Cross(v1, v2)) / Length(v1);//如果不带fabs 得到的是有向距离}//点p到线段ab的距离double DistanceToSegment(Point p, Point a, Point b){    if(a == b) return Length(p-a);    Vector 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p在直线ab上的投影Point GetLineProjection(Point p, Point a, Point b){    Vector v = b-a;    return a + v*(Dot(v, p-a) / Dot(v, v));}//点段相交判定bool SegmentItersection(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;}//点在线段上bool OnSegment(Point p, Point a1, Point a2){    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;}//多变形面积double PolygonArea(Point* p, int n){    double ret = 0;    FF(i, 1, n-1) ret += Cross(p[i]-p[0], p[i+1]-p[0]);    return ret/2;}Point read_point(){    Point a;    scanf("%lf%lf", &a.x, &a.y);    return a;}double torad(double d)//角度转弧度{    return d/180 *acos(-1);}int ConvexHull(Point *p, int n, Point* ch)//凸包{    sort(p, p+n);    int m = 0;    REP(i, n)    {        while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;        ch[m++] = p[i];    }    int k = m;    FD(i, n-2, 0)    {        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;}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;}const int maxn = 10001;int n, T;Point p[maxn], ch[maxn];int main(){    scanf("%d", &T);    FF(kase, 1, T+1)    {        scanf("%d", &n);        double X = 0, Y = 0, ans = 1e10;        REP(i, n) p[i] = read_point(), X += p[i].x, Y += p[i].y;        int m = ConvexHull(p, n, ch);        ch[m] = ch[0];        REP(i, m)        {            double a, b, c;            getLineABC(ch[i], ch[i+1], a, b, c);            ans = min(ans, fabs(a*X+b*Y+c*n)/(sqrt(a*a+b*b)));        }        printf("Case #%d: %.3f\n", kase, n > 2 ? ans/n : 0);    }    return 0;}