题意：凸包上一个点(p)，使得(p)和点(0,1)组成的三角形面积最小
用叉积来求：
(p,i,i+1)组成的三角形面积为： （( imes)为叉积）
((p_p-i) imes (p_p-p_{i+1})Rightarrow)
((x_p-x_i,y_p-y_i) imes(x_p-x_{i+1},y_p-y_{i+1})Rightarrow)
((x_p-x_i)(y_p-y_{i+1})-(y_p-y_i)(x_p-x_{i+1})Rightarrow)
(x_py_p-x_py_{i+1}-x_iy_p+x_iy_{i+1}-x_py_p+x_{i+1}y_p+x_py_i-x_{i+1}y_iRightarrow)
(x_p(y_i-y_{i+1})+y_p(x_{i+1}-x_i)+(x_iy_{i+1}-x_{i+1}y_i))
要求点(p)和点(0,1)组成的三角形面积最小，即：
(x_p(y_0-y_1)+y_p(x_1-x_0)+(x_0y_1-x_1y_0)<x_p(y_i-y_{i+1})+y_p(x_{i+1}-x_i)+(x_iy_{i+1}-x_{i+1}y_i)Rightarrow)
(x_p(y_0-y_1-y_i+y_{i+1})+y_p(x_1-x_0-x_{i+1}+x_i)+(x_0y_1-x_1y_0-x_iy_{i+1}+x_{i+1}y_i)<0)
可以发现，方程为(ax+by+c<0)的形式，可以求出(n)个方程，和原凸多边形求一下半平面交，交出来的面积与原多边形面积的比值即为答案
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<bitset>
#include<sstream>
#include<cstdlib>
#define QAQ int
#define TAT long long
#define OwO bool
#define ORZ double
#define Ug unsigned
#define F(i,j,n) for(QAQ i=j;i<=n;++i)
#define E(i,j,n) for(QAQ i=j;i>=n;--i)
#define MES(i,j) memset(i,j,sizeof(i))
#define MEC(i,j) memcpy(i,j,sizeof(j))
using namespace std;
const QAQ N=300005;
const ORZ eps=1e-8;
QAQ sign(ORZ x){return fabs(x)<=eps ? 0 : (x>0 ? 1 : -1);}
QAQ n;
struct Point {
ORZ x,y;
Point(){}
Point(ORZ X,ORZ Y){x=X;y=Y;}
friend Point operator + (Point a,Point b){
return Point(a.x+b.x,a.y+b.y);
}
friend Point operator - (Point a,Point b){
return Point(a.x-b.x,a.y-b.y);
}
friend Point operator * (Point a,ORZ k){
return Point(a.x*k,a.y*k);
}
friend ORZ operator * (Point a,Point b){
return a.x*b.x+a.y*b.y;
}
friend ORZ operator ^ (Point a,Point b){
return a.x*b.y-a.y*b.x;
}
}p[N];
struct Line{
Point p,v;
ORZ poa;
Line(){}
Line(Point a,Point b){
p=a;v=b;
poa=atan2(b.y,b.x);
}
friend OwO operator < (Line a,Line b){
return sign(a.poa-b.poa)==0 ? sign((a.v) ^ (b.p-a.p)) >0 : sign(a.poa-b.poa)<0;
}
}a[N],q[N];
QAQ js,head,tail,cnt;
ORZ s1,s2;
Point inter(Line a,Line b){
Point u=a.p-b.p;
ORZ k=(b.v^u)/(a.v^b.v);
return a.p+a.v*k;
}
OwO pd(Line a,Point b){
return sign(a.v^(b-a.p))>=0;
}
void Half_Plane(){
sort(a+1,a+js+1);
cnt=1;
F(i,2,js) if(sign(a[i].poa-a[cnt].poa)>0) a[++cnt]=a[i];
head=1;tail=0;
q[++tail]=a[1];q[++tail]=a[2];
F(i,3,cnt){
while(head<tail&&pd(a[i],inter(q[tail-1],q[tail]))) tail--;
while(head<tail&&pd(a[i],inter(q[head+1],q[head]))) head++;
q[++tail]=a[i];
}
while(head<tail&&pd(q[head],inter(q[tail-1],q[tail]))) tail--;
F(i,head,tail-1) p[i]=inter(q[i],q[i+1]);
p[tail]=inter(q[tail],q[head]);
F(i,head,tail-1) s2+=(p[i]^(p[i+1]-p[i]));
s2+=(p[tail]^(p[head]-p[tail]));
}
QAQ main(){
scanf("%d",&n);
F(i,0,n-1) scanf("%lf%lf",&p[i].x,&p[i].y);
p[n]=p[0];
F(i,0,n-1) {
a[++js]=Line(p[i+1],p[i]-p[i+1]);
s1+=(p[i]^(p[i+1]-p[i]));
}
F(i,1,n-1){
ORZ A=p[i+1].x-p[i].x-p[1].x+p[0].x;
ORZ B=p[i+1].y-p[i].y-p[1].y+p[0].y;
ORZ C=-(p[i]^(p[i+1]-p[i]))+(p[0]^(p[1]-p[0]));
if(sign(A)!=0) a[++js]=Line(Point(0,C/A),Point(-A,-B));
else if(sign(B)!=0) a[++js]=Line(Point(-C/B,0),Point(0,-B));
}
Half_Plane();
printf("%.4lf
",fabs(s2/s1));
return 0;
}