luogu4166 最大土地面积 (旋转卡壳)

首先这样的点一定在凸包上

然后旋转卡壳就可以

具体来说，枚举对角线的一个端点，另一个端点在凸包上转，剩下两个点就是一个叉积最大一个最小，而这两个点也是跟着转的

所以是$O(N^2)$

#include<bits/stdc++.h>
#include<tr1/unordered_map>
#define CLR(a,x) memset(a,x,sizeof(a))
#define MP make_pair
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pa;
const int maxn=2005;

inline ll rd(){
    ll x=0;char c=getchar();int neg=1;
    while(c<'0'||c>'9'){if(c=='-') neg=-1;c=getchar();}
    while(c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
    return x*neg;
}

struct Vec{
    double x,y;
    Vec(double a=0,double b=0){x=a,y=b;}
}p[maxn],stk[maxn];
inline Vec operator -(Vec a,Vec b){return Vec(a.x-b.x,a.y-b.y);}
inline double operator *(Vec a,Vec b){return a.x*b.y-a.y*b.x;}
int N,sh;

inline bool cmp(Vec a,Vec b){
    double x=a*b;
    return x?x>0:fabs(a.x)<fabs(b.x);
}

inline int ne(int x,int y=1){return x+y>sh?x+y-sh:x+y;}

int main(){
    //freopen("","r",stdin);
    N=rd();
    for(int i=1;i<=N;i++){
        scanf("%lf%lf",&p[i].x,&p[i].y);
        if(p[i].x<p[1].x||(p[i].x==p[1].x&&p[i].y<p[1].y)) swap(p[i],p[1]);
    }
    for(int i=N;i;i--) p[i]=p[i]-p[1];
    sort(p+1,p+N+1,cmp);
    for(int i=1;i<=N;i++){
        while(sh>1&&(stk[sh]-stk[sh-1])*(p[i]-stk[sh-1])<=0) sh--;
        stk[++sh]=p[i]; 
    }
    double ans=0;
    for(int i=1;i<=sh;i++){
        int a=ne(i,1),b=ne(i,2);
        for(int j=ne(i,2);ne(j)!=i;j=ne(j)){
            while((stk[ne(a)]-stk[i])*(stk[j]-stk[i])>=(stk[a]-stk[i])*(stk[j]-stk[i])) a=ne(a);
            while((stk[j]-stk[i])*(stk[ne(b)]-stk[i])>=(stk[j]-stk[i])*(stk[b]-stk[i])) b=ne(b);
            ans=max(ans,((stk[a]-stk[i])*(stk[j]-stk[i])+(stk[j]-stk[i])*(stk[b]-stk[i]))/2);
        }
    }
    printf("%.3lf\n",ans);
    return 0;
}