版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/leekerian/article/details/82154010
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=11111;
const double eps=1e-8;
int sgn(double x)
{
if(fabs(x)<eps)
return 0;
if(x<0)
return -1;
else
return 1;
}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y)
{
x=_x;
y=_y;
}
Point operator -(const Point &b)const
{
return Point(x-b.x,y-b.y);
}
Point operator +(const Point &b)const
{
return Point(x+b.x,y+b.y);
}
double operator ^(const Point &b)const
{
return x*b.y-y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x+y*b.y;
}
};
Point p[maxn];
int n;
int Stack[maxn];
int top;
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
bool cmp(Point a,Point b)
{
int ans=sgn((a-p[0])^(b-p[0]));
if(ans==1)
return true;
else if(ans==0)
return dist(a,p[0])<dist(b,p[0]);
else
return false;
}
void graham()
{
for(int i=0;i<n;i++)
{
if(p[i].x<p[0].x||(p[i].x==p[0].x&&p[i].y<p[0].y))
swap(p[i],p[0]);
}
sort(p+1,p+n,cmp);
Stack[0]=0;
Stack[1]=1;
top=2;
for(int i=2;i<n;i++)
{
while(top>1&&sgn((p[Stack[top-1]]-p[Stack[top-2]])^(p[i]-p[Stack[top-2]]))<=0)
top--;
Stack[top++]=i;
}
}
Point p1[maxn];
Point p_1,p_2,p_3;
void r_a()
{
int l=1;
int l1=1;
double res=0;
for(int i=0;i<top;i++)
{
while(sgn(((p1[l1]-p1[i])^(p1[l]-p1[i]))-((p1[l1]-p1[i])^(p1[(l+1)%top]-p1[i])))<0)
l=(l+1)%top;
while(sgn(((p1[l1]-p1[i])^(p1[l]-p1[i]))-((p1[(l1+1)%top]-p1[i])^(p1[l]-p1[i])))<0)
l1=(l1+1)%top;
if(sgn(((p1[l1]-p1[i])^(p1[l]-p1[i]))-res)>=0)
{
res=((p1[l1]-p1[i])^(p1[l]-p1[i]));
p_1=p1[i];
p_2=p1[l1];
p_3=p1[l];
}
}
//cout<<res<<endl;
}
int main()
{
long long s;
scanf("%d %lld",&n,&s);
for(int i=0;i<n;i++)
{
scanf("%lf %lf",&p[i].x,&p[i].y);
}
graham();
for(int i=0;i<top;i++)
{
p1[i]=p[Stack[i]];
//cout<<p1[i].x<<" "<<p1[i].y<<endl;
}
r_a();
//cout<<p_1.x<<" "<<p_1.y<<endl;
//cout<<p_2.x<<" "<<p_2.y<<endl;
//cout<<p_3.x<<" "<<p_3.y<<endl;
Point ans1;
Point ans2;
Point ans3;
ans1=p_3-p_1+p_2;
ans2=p_1-p_3+p_2;
ans3=p_3-p_2+p_1;
printf("%.0lf %.0lf\n",ans1.x,ans1.y);
printf("%.0lf %.0lf\n",ans2.x,ans2.y);
printf("%.0lf %.0lf\n",ans3.x,ans3.y);
return 0;
}