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0
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;
}

简单几何 cf198 B. Maximal Area Quadrilateral,布布扣,bubuko.com

简单几何 cf198 B. Maximal Area Quadrilateral

原文：http://blog.csdn.net/guognib/article/details/20792557