mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
80 lines
1.9 KiB
C++
80 lines
1.9 KiB
C++
|
#include<cstdio>
|
||
|
#include<iostream>
|
||
|
#include<utility>
|
||
|
#include<string>
|
||
|
#include<algorithm>
|
||
|
#include<cmath>
|
||
|
using namespace std;
|
||
|
const int M=15;
|
||
|
const double inf=1e250;
|
||
|
struct point
|
||
|
{
|
||
|
double x,y;
|
||
|
bool operator < (const point a)const
|
||
|
{
|
||
|
if(y==a.y)
|
||
|
return x<a.x;
|
||
|
return y<a.y;
|
||
|
}
|
||
|
}p[M],stac[M];
|
||
|
int n,top;
|
||
|
double dis(point a,point b)
|
||
|
{
|
||
|
return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));
|
||
|
}
|
||
|
double xmult(double x1,double y1,double x2,double y2)
|
||
|
{
|
||
|
return x1*y2-x2*y1;
|
||
|
}
|
||
|
double cross(point p0,point p1,point p2)
|
||
|
{
|
||
|
double x1=p1.x-p0.x,y1=p1.y-p0.y;
|
||
|
double x2=p2.x-p0.x,y2=p2.y-p0.y;
|
||
|
return xmult(x1,y1,x2,y2);
|
||
|
}
|
||
|
void covex_hull()
|
||
|
{
|
||
|
sort(p,p+n);
|
||
|
stac[0]=p[0];
|
||
|
stac[1]=p[1];
|
||
|
top=1;
|
||
|
for(int i=2;i<n;i++)
|
||
|
{
|
||
|
while(top>0 && cross(stac[top],p[i],stac[top-1])<=0)
|
||
|
top--;
|
||
|
stac[++top]=p[i];
|
||
|
}
|
||
|
int temp=top;
|
||
|
for(int i=n-2;i>=0;i--)
|
||
|
{
|
||
|
while(top>temp && cross(stac[top],p[i],stac[top-1])<=0)
|
||
|
top--;
|
||
|
stac[++top]=p[i];
|
||
|
}
|
||
|
}
|
||
|
int main()
|
||
|
{
|
||
|
while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p[0].x,&p[0].y,&p[1].x,&p[1].y,&p[2].x,&p[2].y,&p[3].x,&p[3].y)==8)
|
||
|
{
|
||
|
if(p[0].x==-1 && p[0].y==-1 && p[1].x==-1 && p[1].y==-1 && p[2].x==-1 && p[2].y==-1 && p[3].x==-1 && p[3].y==-1)
|
||
|
break;
|
||
|
n=4;
|
||
|
covex_hull();
|
||
|
double ans;
|
||
|
if(top==4)
|
||
|
{
|
||
|
ans=dis(stac[0],stac[2])+dis(stac[1],stac[3]);
|
||
|
}
|
||
|
else
|
||
|
{
|
||
|
ans=inf;
|
||
|
ans=min(ans,dis(p[0],p[1])+dis(p[0],p[2])+dis(p[0],p[3]));
|
||
|
ans=min(ans,dis(p[1],p[0])+dis(p[1],p[2])+dis(p[1],p[3]));
|
||
|
ans=min(ans,dis(p[2],p[1])+dis(p[2],p[0])+dis(p[2],p[3]));
|
||
|
ans=min(ans,dis(p[3],p[1])+dis(p[3],p[2])+dis(p[3],p[0]));
|
||
|
}
|
||
|
printf("%.4lf\n",ans);
|
||
|
}
|
||
|
return 0;
|
||
|
}
|