hdu 3694 10 福州 现场 E - Fermat Point in Quadrangle 费马点 计算几何 难度:1
Alice and Bob are learning geometry. Recently they are studying about the Fermat Point.
Alice: I wonder whether there is a similar point for quadrangle.
Bob: I think there must exist one.
Alice: Then how to know where it is? How to prove?
Bob: I don’t know. Wait… the point may hold the similar property as the case in triangle.
Alice: It sounds reasonable. Why not use our computer to solve the problem? Find the Fermat point, and then verify your assumption.
Bob: A good idea.
So they ask you, the best programmer, to solve it. Find the Fermat point for a quadrangle, i.e. find a point such that the total distance from the four vertices of the quadrangle to that point is the minimum.
Input
Each test case is a single line which contains eight float numbers, and it is formatted as below:
x 1 y 1 x 2 y 2 x 3 y 3 x 4 y 4
x i, y i are the x- and y-coordinates of the ith vertices of a quadrangle. They are float numbers and satisfy 0 ≤ x i ≤ 1000 and 0 ≤ y i ≤ 1000 (i = 1, …, 4).
The input is ended by eight -1.
Output
Sample Input
1 1 1 1 1 1 1 1
-1 -1 -1 -1 -1 -1 -1 -1
四边形费马点
平面四边形费马点证明图形
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std; const double eps=1e-10; double add(double a,double b)
{
if(abs(a+b)<eps*(abs(a)+abs(b))) return 0;
return a+b;
} struct point
{
double x,y;
point () {}
point (double x,double y) : x(x),y(y){ }
point operator + (point p)
{
return point (add(x,p.x),add(y,p.y));
}
point operator - (point p)
{
return point (add(x,-p.x),add(y,-p.y));
}
point operator * (double d)
{
return point (x*d,y*d);
}
double dot(point p)
{
return add(x*p.x,y*p.y);
}
double det(point p)
{
return add(x*p.y,-y*p.x);
}
}; bool on_seg(point p1,point p2,point q)
{
return (p1-q).det(p2-q)==0&&(p1-q).dot(p2-q)<=0;
} point intersection(point p1,point p2,point q1,point q2)
{
return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));
} bool cmp_x(const point&p,const point& q)
{
if(p.x!=q.x) return p.x<q.x;
return p.y<q.y;
} vector<point> convex_hull(point*ps,int n)
{
sort(ps,ps+n,cmp_x);
//for(int i=0;i<n;i++) printf("x=%.f %.f")
int k=0;
vector<point> qs(n*2);
for(int i=0;i<n;i++){
while(k>1&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0) k--;
qs[k++]=ps[i];
}
for(int i=n-2,t=k;i>=0;i--){
while(k>t&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0) k--;
qs[k++]=ps[i];
}
qs.resize(k-1);
return qs;
} double dis(point p1,point p2)
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
bool equ(point p1,point p2)
{
if(fabs(p1.x-p2.x)<eps&&fabs(p1.y-p2.y)<eps)
return true;
return false;
}
int main()
{
point p[10];
for(int i=0;i<4;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
while(p[0].x!=-1&&p[0].y!=-1)
{
vector <point> m;
double minn=100000000,d;
m=convex_hull(p,4);//检查是否四边形
if(m.size()==4)//如果是四边形则加入对角线交点考虑
minn=dis(m[1],m[3])+dis(m[0],m[2]);
for(int i=0;i<4;i++)
{
d=0;
for(int j=0;j<4;j++)
d+=dis(p[i],p[j]);
minn=min(minn,d);
}
printf("%.4f\n",minn);
for(int i=0;i<4;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
}
return 0;
}
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