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hdu 3694 10 福州 现场 E - Fermat Point in Quadrangle

2024-08-01 09:26:43   219人阅读

In geometry the Fermat point of a triangle, also called Torricelli point, is a point such that the total distance from the three vertices of the triangle to the point is the minimum. It is so named because this problem is first raised by Fermat in a private letter. In the following picture, P 0 is the Fermat point. You may have already known the property that: 

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

The input contains no more than 1000 test cases. 

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

For each test case, find the Fermat point, and output the total distance from the four vertices to that point. The result should be rounded to four digits after the decimal point.
 

Sample Input

0 0 1 1 1 0 0 11 1 1 1 1 1 1 1-1 -1 -1 -1 -1 -1 -1 -1
 
取四个点其中一个点或者四个点两两连线的交点
#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);        if(m.size()==4)            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;}

　　

hdu 3694 10 福州 现场 E - Fermat Point in Quadrangle


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