1 #include <cstdio>  2 #include <algorithm>  3 #include <cmath>  4 #include <vector>  5 using namespace std;  6   7 const int INF = (1<<30)-1;  8   9 //lrj计算几何模板 10 struct Point 11 { 12     int x, y; 13     Point(double x=0, double y=0) :x(x),y(y) {} 14 }; 15 typedef Point Vector; 16  17 Point read_point(void) 18 { 19     double x, y; 20     scanf("%lf%lf", &x, &y); 21     return Point(x, y); 22 } 23  24 const double EPS = 1e-10; 25  26 //向量+向量=向量 点+向量=点 27 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); } 28  29 //向量-向量=向量 点-点=向量 30 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); } 31  32 //向量*数=向量 33 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); } 34  35 //向量/数=向量 36 Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); } 37  38 bool operator < (const Point& a, const Point& b) 39 { return a.x < b.x || (a.x == b.x && a.y < b.y); } 40  41 int dcmp(double x) 42 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; } 43  44 bool operator == (const Point& a, const Point& b) 45 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; } 46  47 /**********************基本运算**********************/ 48  49 //点积 50 double Dot(Vector A, Vector B) 51 { return A.x*B.x + A.y*B.y; } 52 //向量的模 53 double Length(Vector A)    { return sqrt(Dot(A, A)); } 54  55 //向量的夹角，返回值为弧度 56 double Angle(Vector A, Vector B) 57 { return acos(Dot(A, B) / Length(A) / Length(B)); } 58  59 //叉积 60 double Cross(Vector A, Vector B) 61 { return A.x*B.y - A.y*B.x; } 62  63 //向量AB叉乘AC的有向面积 64 double Area2(Point A, Point B, Point C) 65 { return Cross(B-A, C-A); } 66  67 //向量A旋转rad弧度 68 Vector VRotate(Vector A, double rad) 69 { 70     return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); 71 } 72  73 //将B点绕A点旋转rad弧度 74 Point PRotate(Point A, Point B, double rad) 75 { 76     return A + VRotate(B-A, rad); 77 } 78  79 //求向量A向左旋转90°的单位法向量,调用前确保A不是零向量 80 Vector Normal(Vector A) 81 { 82     double l = Length(A); 83     return Vector(-A.y/l, A.x/l); 84 } 85  86 /**********************点和直线**********************/ 87  88 //求直线P + tv 和 Q + tw的交点，调用前要确保两条直线有唯一交点 89 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) 90 { 91     Vector u = P - Q; 92     double t = Cross(w, u) / Cross(v, w); 93     return P + v*t; 94 }//在精度要求极高的情况下，可以自定义分数类 95  96 //P点到直线AB的距离 97 double DistanceToLine(Point P, Point A, Point B) 98 { 99     Vector v1 = B - A, v2 = P - A;100     return fabs(Cross(v1, v2)) / Length(v1);    //不加绝对值是有向距离101 }102 103 //点到线段的距离104 double DistanceToSegment(Point P, Point A, Point B)105 {106     if(A == B)    return Length(P - A);107     Vector v1 = B - A, v2 = P - A, v3 = P - B;108     if(dcmp(Dot(v1, v2)) < 0)    return Length(v2);109     else if(dcmp(Dot(v1, v3)) > 0)    return Length(v3);110     else return fabs(Cross(v1, v2)) / Length(v1);111 }112 113 //点在直线上的射影114 Point GetLineProjection(Point P, Point A, Point B)115 {116     Vector v = B - A;117     return A + v * (Dot(v, P - A) / Dot(v, v));118 }119 120 //线段“规范”相交判定121 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)122 {123     double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);124     double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);125     return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;126 }127 128 //判断点是否在线段上129 bool OnSegment(Point P, Point a1, Point a2)130 {131     Vector v1 = a1 - P, v2 = a2 - P;132     return dcmp(Cross(v1, v2)) == 0 && dcmp(Dot(v1, v2)) < 0;133 }134 135 //求多边形面积136 double PolygonArea(Point* P, int n)137 {138     double ans = 0.0;139     for(int i = 1; i < n - 1; ++i){140         double tmp = Cross(P[i]-P[0], P[i+1]-P[0]);141         tmp = fabs(tmp);142         //printf("tmp = %lf\n",tmp);143         ans += tmp;144     }145     return ans/2;146 }147 148 Point p[5],a[505];149 int n;150 151 void solve(){152     double ans = 0.0;153     for(int i = 1;i <= n;i++){154         for(int j = i+1;j <= n;j++){155             double minn = -1.0*INF;156             double maxx = -1.0*INF;157             for(int k = 1;k <= n;k++){158                 if(k == i || k == j) continue;159 //                printf("i = %d j = %d k = %d ",i,j,k);160                 double tmp = Cross(a[i]-a[j], a[k]-a[j]);161                 if(tmp < 0) {162                     tmp = -tmp;163                     minn = max(tmp,minn);164                 }165                 else maxx = max(maxx,tmp);166             }167             ans = max(ans,(minn+maxx)/2.0);168         }169     }170     printf("%.12lf\n",ans);171 }172 173 int main(){174     while(scanf("%d",&n) != EOF){175         for(int i = 1;i <= n;i++){176             scanf("%d %d",&a[i].x,&a[i].y);177         }178         solve();179     }180     return 0;181 }