简单几何 cf198 B. Maximal Area Quadrilateral

cf198 B. Maximal Area Quadrilateral

300个点，取4个点组成一个四边形，求四边形最大面积？

解法：枚举四边形的对角线，再求所有其它点的叉积，取最大最小值，再绝对值相加即可

（几何向量解法中常见的和面积相关的就是三角形的有向面积，即叉积！！！）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <cstdlib>#include <fstream>#include <vector>#include <set>using namespace std;const double eps = 1e-12;struct Point {    double x, y;    Point(double x = 0, double y = 0) : x(x), y(y) {}    void In()    {        scanf("%lf%lf", &x, &y);    }};typedef Point Vector;Vector operator+(Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }Vector operator-(Point A, Point 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;    else 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; }///(x, y) 极角atan2(y, x);单位：弧度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 Angle(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(Point A, Point B, Point C) { return Cross(B - A, C - A); }///有向Vector Rotate(Vector A, double rad) ///zhengming{    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));}Vector Normal(Vector A)///法向量{    double L = Length(A);    return Vector(-A.y / L, A.x / L);}///直线表示P=P0 + tV; P = A + (B-A)t; t 为参数///直线相交,求之前保证P + tv 和 Q + tv 有唯一的交点。当且仅当Cross(V, W)非0Point 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;}///点到直线距离double DistanceToLine(Point P, Point A, Point B)///!!!!!!!!!Cross{    Vector v1 = B - A, v2 = P - A;    return fabs(Cross(v1, v2)) / Length(v1);///不取绝对值为有向距离}///点到线段距离double DistanceToSegment(Point P, Point A, Point B)///Dot/Cross{    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);///判断点在线段的位置和 Dot、Cross的四个象限的图联系    else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);    else return fabs(Cross(v1, v2)) / Length(v1);///不取绝对值为有向距离}///点到直线的投影，Dot的分配率Point GetLineProjection(Point P, Point A, Point B){    Vector v = B - A;    return A + v * (Dot(v, P - A) / Dot(v, v));}///线段相交3种情况///1.线段相交判定（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);    double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;}///2.c1 c2 都为0，想段共线(不平行)，可能部分重合///3.c1 c2 不都为0，///点在线上的判断bool Onsegment(Point p, Point a1, Point a2){    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;}double PolygonArea(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;}Point t[320];int main(){    int n;    cin >> n;    double ans = 0.0;    for (int i = 0; i < n; i++)    {        t[i].In();    }    for (int i = 0; i < n; i++)    {        for (int j = 0; j < n; j++)        {            if (j == i) continue;///保证枚举所有的对角线            Vector p1 = t[i] - t[j];            double area_max = 0.0;            double area_min = 0.0;            for (int r = 0; r < n; r++)            {                if (i == r || j == r) continue;///保证枚举所有的对角线                Vector p2 = t[i] - t[r];                double now = Cross(p1, p2);                area_max = max(now, area_max);                area_min = min(now, area_min);            }            if (area_max == 0 || area_min == 0) continue;///保证两点在对角线的两边            ans = max(ans, fabs(area_max) + fabs(area_min));        }    }    printf("%.10lf\n", ans / 2.0);    return 0;}