Codeforces Round #358 (Div. 2) E 计算几何 旋转卡壳求最大三角形面积

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E. Alyona and Triangles

time limit per test3 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

You are given n points with integer coordinates on the plane. Points are given in a way such that there is no triangle, formed by any three of these n points, which area exceeds S.

Alyona tried to construct a triangle with integer coordinates, which contains all n points and which area doesn't exceed 4S, but, by obvious reason, had no success in that. Please help Alyona construct such triangle. Please note that vertices of resulting triangle are not necessarily chosen from n given points.

Input
In the first line of the input two integers n and S (3 ≤ n ≤ 5000, 1 ≤ S ≤ 1018) are given — the number of points given and the upper bound value of any triangle's area, formed by any three of given n points.

The next n lines describes given points: ith of them consists of two integers xi and yi ( - 108 ≤ xi, yi ≤ 108) — coordinates of ith point.

It is guaranteed that there is at least one triple of points not lying on the same line.

Output
Print the coordinates of three points — vertices of a triangle which contains all n points and which area doesn't exceed 4S.

Coordinates of every triangle's vertex should be printed on a separate line, every coordinate pair should be separated by a single space. Coordinates should be an integers not exceeding 109 by absolute value.

It is guaranteed that there is at least one desired triangle. If there is more than one answer, print any of them.

Example
input
4 1
0 0
1 0
0 1
1 1
output
-1 0
2 0
0 2
Note

题意：

给出n个点，要求找出一个最大的三角形使得三角形覆盖这n个点。三角形的面积不能超过四倍的点集区域面积。

思路：

旋转卡壳求出凸包内最大的三角形ABC（图片来自卿神）

显然三角形A'B'C'就是答案。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
struct point{
    double x,y;
    point(double x=0,double y=0):x(x),y(y){}
    bool operator < (const point &a)const{
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
};
const double eps=1e-8;
typedef point vec;
vec operator + (point a,point b){
    return vec(a.x+b.x,a.y+b.y);
}
vec operator - (point a,point b){
    return vec(a.x-b.x,a.y-b.y);
}
vec operator * (point a,double t){
    return vec(a.x*t,a.y*t);
}
vec operator / (point a,double t){
    return vec(a.x/t,a.y/t);
}
double cross(vec a,vec b){
    return a.x*b.y-a.y*b.x;
}
int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x<0?-1:1;
}
bool cmp(point a,point b){
    if(dcmp(a.x-b.x)==0) return a.y<b.y;
    return a.x<b.x;
}
point p[5010];
void convexhull(point *s,int &n){
    sort(s,s+n);
    ///sort(s,s+n,cmp);
    int m=0;
    s[n]=s[0];
    for(int i=0;i<n;i++){
        while(m>1 && dcmp(cross(p[m-1]-p[m-2],s[i]-p[m-2]))<=0)
            m--;
        p[m++]=s[i];
    }
    int k=m;
    for(int i=n-2;i>=0;i--){
        while(m>k && dcmp(cross(p[m-1]-p[m-2],s[i]-p[m-2]))<=0)
            m--;
        p[m++]=s[i];
    }
    m--;
    n=m;
    for(int i=0;i<n;i++) s[i]=p[i];
}
point s[5010];
ll S;
int n;
int main(){
    scanf("%d%I64d",&n,&S);
    for(int i=0;i<n;i++) scanf("%lf%lf",&s[i].x,&s[i].y);
    convexhull(s,n);
    s[n]=s[0];
    s[n+1]=s[1];
    s[n+2]=s[2];
    double sum=0.0;
    int ansi=0,ansj=1,ansk=2;
    int j=1,k=2;
    for(int i=0;i<n;i++){
        while(fabs(cross(s[j]-s[i],s[k+1]-s[i]))-fabs(cross(s[j]-s[i],s[k]-s[i]))>=eps)
            k=(k+1)%n;
        while(fabs(cross(s[k]-s[i],s[j+1]-s[i]))-fabs(cross(s[k]-s[i],s[j]-s[i]))>=eps)
            j=(j+1)%n;
        double tmp=fabs(cross(s[j]-s[i],s[k]-s[i]));
        if(dcmp(tmp-sum)>=0){
            ansi=i;
            ansj=j;
            ansk=k;
            sum=tmp;
        }
    }
    point A=s[ansj]+s[ansk]-s[ansi];
    point B=s[ansi]+s[ansk]-s[ansj];
    point C=s[ansi]+s[ansj]-s[ansk];
    printf("%.0f %.0f\n",A.x,A.y);
    printf("%.0f %.0f\n",B.x,B.y);
    printf("%.0f %.0f\n",C.x,C.y);
    return 0;
}