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=;

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

 struct Vec{
     double x,y;
     Vec(double a=,double b=){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>:fabs(a.x)<fabs(b.x);
 }

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

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