#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define error(x) (!(x)?puts("error"):0)#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }const int N=2005;struct dat { double x, y; }a[N], b[N];int n, cnt;double suml[N][N], sumr[N][N];bool cmp(const dat &a, const dat &b) { return a.x<b.x; }double cross(const dat &a, const dat &b, const dat &c) {	static dat x, y;	x.x=a.x-c.x; x.y=a.y-c.y;	y.x=b.x-c.x; y.y=b.y-c.y;	return x.x*y.y-y.x*x.y;}void tb() {	sort(a+1, a+1+n, cmp);	for1(i, 1, n) {		while(cnt>1 && cross(b[cnt], a[i], b[cnt-1])<=0) --cnt;		b[++cnt]=a[i];	}	int k=cnt;	for3(i, n-1, 1) {		while(cnt>k && cross(b[cnt], a[i], b[cnt-1])<=0) --cnt;		b[++cnt]=a[i];	}	if(n>1) --cnt;}double S(const dat &a, const dat &b, const dat &c) { return abs(cross(a, b, c))/2.0; }void work() {	int mid, nxt, j;	for1(i, 1, n) {		mid=(i%n)+1;		for1(t, 1, n-2) {			j=(i+t)%n+1;			nxt=(mid%n)+1;			while(nxt!=j && S(b[i], b[j], b[nxt])>=S(b[i], b[j], b[mid])) { mid=nxt; nxt=(mid%n)+1; }			suml[i][j]=S(b[i], b[j], b[mid]); //printf("%d %d chose:%d: S:%.3f\n", i, j, mid, suml[i][j]);		}		mid=(i-2+n)%n+1;		for1(t, 1, n-2) {			j=(i-2+n-t+n)%n+1;			nxt=(mid-2+n)%n+1;			while(nxt!=j && S(b[i], b[j], b[nxt])>=S(b[i], b[j], b[mid])) { mid=nxt; nxt=((mid-2+n)%n)+1; }			sumr[i][j]=S(b[i], b[j], b[mid]);		}	}}int main() {	read(n);	for1(i, 1, n) scanf("%lf%lf", &a[i].x, &a[i].y);	tb(); n=cnt;	//for1(i, 1, n) printf("%.2f %.2f\n", b[i].x, b[i].y);	work(); double ans=0;	for1(i, 1, n) for1(j, 1, n) if(suml[i][j] && sumr[i][j]) ans=max(ans, suml[i][j]+sumr[i][j]);	printf("%.3f", ans);	return 0;}

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define error(x) (!(x)?puts("error"):0)#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }const int N=2005;struct dat { double x, y; }a[N], b[N];int n, cnt;bool cmp(const dat &a, const dat &b) { return a.x<b.x; }double cross(const dat &a, const dat &b, const dat &c) {	static dat x, y;	x.x=a.x-c.x; x.y=a.y-c.y;	y.x=b.x-c.x; y.y=b.y-c.y;	return x.x*y.y-y.x*x.y;}void tb() {	sort(a+1, a+1+n, cmp);	for1(i, 1, n) {		while(cnt>1 && cross(b[cnt], a[i], b[cnt-1])<=0) --cnt;		b[++cnt]=a[i];	}	int k=cnt;	for3(i, n-1, 1) {		while(cnt>k && cross(b[cnt], a[i], b[cnt-1])<=0) --cnt;		b[++cnt]=a[i];	}	if(n>1) --cnt;}double S(const dat &a, const dat &b, const dat &c) { return abs(cross(a, b, c)); }double work() {	b[cnt+1]=b[1];	int dmid, umid;	double ret=0;	for1(i, 1, n) {		dmid=(i%n)+1, umid=((i+2)%n)+1;		for1(j, i+2, n) {			while((dmid%n)+1!=j && S(b[i], b[j], b[dmid+1])>=S(b[i], b[j], b[dmid])) dmid=(dmid%n)+1;			while((umid%n)+1!=i && S(b[i], b[j], b[umid+1])>=S(b[i], b[j], b[umid])) umid=(umid%n)+1;			ret=max(ret, S(b[i], b[j], b[dmid])+S(b[i], b[j], b[umid]));//printf("%d %d chose:%d: S:%.3f\n", i, j, mid, suml[i][j]);		}	}	return ret/2.0;}int main() {	read(n);	for1(i, 1, n) scanf("%lf%lf", &a[i].x, &a[i].y);	tb(); n=cnt;	//for1(i, 1, n) printf("%.2f %.2f\n", b[i].x, b[i].y);	printf("%.3f", work());	return 0;}

在某块平面土地上有N个点，你可以选择其中的任意四个点，将这片土地围起来，当然，你希望这四个点围成的多边形面积最大。

第1行一个正整数N，接下来N行，每行2个数x,y，表示该点的横坐标和纵坐标。

最大的多边形面积，答案精确到小数点后3位。

数据范围 n<=2000, |x|,|y|<=100000