1069: [SCOI2007]最大土地面积

//貌似本题与旋转卡壳没有太大的关系
//个人觉得致力不错：http://www.cnblogs.com/mjy0724/p/4423905.html
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=1e5+5;
const double eps=1e-10;
struct Vector{
    double x,y;
    Vector(double x=0,double y=0):x(x),y(y){}
}p[N],ch[N];int n;
Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}//向量加法
Vector operator - (Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}//向量减法
Vector operator * (Vector A,double p){return Vector(A.x*p,A.y*p);}//向量乘以标量
Vector operator / (Vector A,double p){return Vector(A.x/p,A.y/p);}//向量除以标量
bool operator < (const Vector &a,const Vector &b){//点的坐标排序
    return a.x==b.x?a.y<b.y:a.x<b.x;
}
int dcmp(double x){//三态函数，减少精度问题
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;//忘写return WA*1 
}
bool operator ==(const Vector &a,const Vector &b){//判断同一个点
    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;}//向量点积
double Length(Vector A){return sqrt(Dot(A,A));} //向量长度，点积
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}//向量转角，逆时针，点积
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}//向量叉积
double Area2(Vector A,Vector B,Vector C){return Cross(B-A,C-A);}//三角形有向面积的两倍 
//计算凸包，输入点数组p,不重复个数cnt,输出点数组ch。函数返回凸包顶点个数。
//如果不希望在凸包的边上有输入点，把两个 <=改成 < 
int ConvexHull(){
    sort(p,p+n);
    int m=0;ch[0]=p[0];
    for(int i=1;i<n;i++){
        while(m>0&&Cross(ch[m]-ch[m-1],p[i]-ch[m-1])<=0) m--;
        ch[++m]=p[i];
    }
    int k=m;
    for(int i=n-2;i>=0;i--){
        while(m>k&&Cross(ch[m]-ch[m-1],p[i]-ch[m-1])<=0) m--;
        ch[++m]=p[i];
    }
    return m+=(n==1);
}
double maxarea(int n,Vector *p){
    if(n<=2) return 0.0;
    if(n==3) return fabs(Area2(p[0],p[1],p[2]));
    static int nxt[N];
    nxt[n-1]=0;
    for(int i=0;i<n-1;i++) nxt[i]=i+1;
    double rt=0.0;
    for(int i=0;i<n;i++){
        for(int j=nxt[nxt[i]],a=nxt[i],b=nxt[j];nxt[j]!=i;j=nxt[j]){
            while(Area2(p[i],p[a],p[j])<Area2(p[i],p[nxt[a]],p[j])) a=nxt[a];
            while(Area2(p[j],p[b],p[i])<Area2(p[j],p[nxt[b]],p[i])) b=nxt[b];
            double ra=Area2(p[i],p[a],p[j]);
            double rb=Area2(p[j],p[b],p[i]);
            rt=max(rt,ra+rb);
        }
    }
    return rt/2.0;
}
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
    int c=ConvexHull();
    printf("%.3lf\n",maxarea(c,ch));
    return 0;
}