[SCOI2007]最大土地面积（旋转卡壳）

首先，最大四边形的四个点一定在凸包上

所以先求凸包

有个结论，若是随机数据，凸包包括的点大约是\(\log_2n\)个

然鹅，此题绝对不会这么轻松，若\(O(n^4)\)枚举，只有50分

所以还是要想正解

旋转卡壳是继承上一个点枚举，所以枚举对角线上的两点，通过旋转卡壳找剩余两点

复杂度\(O(n^2)\)

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<bitset>
#include<sstream>
#include<cstdlib>
#define QAQ int
#define TAT long long
#define OwO bool
#define ORZ double
#define F(i,j,n) for(QAQ i=j;i<=n;++i)
#define E(i,j,n) for(QAQ i=j;i>=n;--i)
#define MES(i,j) memset(i,j,sizeof(i))
#define MEC(i,j) memcpy(i,j,sizeof(j))

using namespace std;
const QAQ N=200005;
const ORZ eps=1e-8;

QAQ n;
struct Point{
	ORZ x,y;
	friend Point operator + (Point a,Point b){
		Point t;
		t.x=a.x+b.x;t.y=a.y+b.y;
		return t;
	}
	friend Point operator - (Point a,Point b){
		Point t;
		t.x=a.x-b.x;t.y=a.y-b.y;
		return t;
	}
	friend ORZ operator ^ (Point a,Point b){
		return a.x*b.y-a.y*b.x;
	}
	friend ORZ operator * (Point a,Point b){
		return a.x*b.x+a.y*b.y;
	}
}a[N],s[N];
QAQ top;
ORZ ans;

QAQ sign(ORZ x){
	return fabs(x)<=eps ? 0 : (x>0 ? 1 : -1);
}

ORZ dis(Point i,Point j){
	return (i.x-j.x)*(i.x-j.x)+(i.y-j.y)*(i.y-j.y);
}

OwO comp(Point i,Point j){
	ORZ x=(i-a[1])^(j-a[1]);
	return x>0||x==0&&dis(a[1],i)<dis(a[1],j);
}

void Graham(){
	QAQ k=1;
	F(i,2,n) if(a[i].y<a[k].y||(a[i].y==a[k].y&&a[i].x<a[k].x)) k=i;
	swap(a[k],a[1]);
	sort(a+2,a+n+1,comp);
	s[++top]=a[1];s[++top]=a[2];
	F(i,3,n){
		while(top>=2&&sign((s[top]-s[top-1]) ^ (a[i]-s[top-1]))<=0) top--; //"<=0" 别忘"="
		s[++top]=a[i];
	}
}

ORZ cal(Point i,Point j,Point k,Point l){
	return (((k-i)^(j-i))+((l-i)^(k-i)))/2.0;
}

ORZ work(){
	ORZ ans=0;
	s[top+1]=a[1];
	F(i,1,top){
		QAQ a=i%top+1,b=(i+2)%top+1;
		F(j,i+2,top){
			while(a%top+1!=j&&(((s[a]-s[i])^(s[j]-s[i])))<(((s[a+1]-s[i])^(s[j]-s[i])))) (a%=top)+=1;
			while(b%top+1!=j&&(((s[j]-s[i])^(s[b]-s[i])))<(((s[j]-s[i])^(s[b+1]-s[i])))) (b%=top)+=1;
			//注意叉积的前后向量顺序
			ans=max(ans,fabs(((s[a]-s[i])^(s[j]-s[i]))+((s[j]-s[i])^(s[b]-s[i]))));
		}
	}
	return ans;
}

QAQ main(){
	scanf("%d",&n);
	F(i,1,n) scanf("%lf%lf",&a[i].x,&a[i].y);
	Graham();
	printf("%.3lf\n",work());
	return 0;
}