先把位于凸包的点求出，然后n^2枚举每两个点x,y，接着左右边找个离线最远的点。

可以知道，当x不变y单调递增时，两边距离最远的两点也在单调递增。

于是可以使用旋转卡壳。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #define rep(i, l, r) for(int i=l; i<=r; i++)#define down(i, l, r) for(int i=l; i>=r; i--)#define maxn 2009#define pi acos(-1)#define eps 1e-11using namespace std;inline int read(){ int x=0, f=1; char ch=getchar(); while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();} while (isdigit(ch)) x=x*10+ch-'0', ch=getchar(); return x*f;}struct P{double x, y;} p[maxn], s[maxn];P operator - (P a, P b){return (P){a.x-b.x, a.y-b.y};}double operator * (P a, P b){return a.x*b.y-a.y*b.x;}double dis(P a, P b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}bool cmp(P a, P b){ double t=(a-p[1])*(b-p[1]); if (t==0) return dis(a, p[1])
            
             1 && (p[i]-s[top-1])*(s[top]-s[top-1])<=0) top--; s[++top]=p[i]; } s[top+1]=p[1]; int a, b; rep(x, 1, top) { a=x%top+1; b=(x+2)%top+1; rep(y, x+2, top) { while (a%top+1!=y && (s[y]-s[x])*(s[a+1]-s[x])>(s[y]-s[x])*(s[a]-s[x])) a=a%top+1; while (b%top+1!=x && (s[b+1]-s[x])*(s[y]-s[x])>(s[b]-s[x])*(s[y]-s[x])) b=b%top+1; ans=max((s[y]-s[x])*(s[a]-s[x])+(s[b]-s[x])*(s[y]-s[x]), ans); } } printf("%.3lf\n", ans/2); return 0;}