2010 Asia Fuzhou Regional Contest HDOJ 3694 Fermat Point in Quadrangle

链接：http://acm.hdu.edu.cn/showproblem.php?pid=3694

题目:

Fermat Point in Quadrangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1851    Accepted Submission(s): 319

Problem Description

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, P₀ 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:

x₁ y₁ x₂ y₂ x₃ y₃ x₄ y₄

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

 

Sample Output

2.82840.0000

 

题意：

求四边形的费马点到四个顶点的距离之和。

解题思路：

费马点（到各顶点距离和最小的点）的基本应用。结论：对一个四边形，如果是凹四边形，费马点为凹点；如果是凸四边形，费马点为对角线的交点。

代码：

#include <cstdio>#include <cmath>#include <algorithm>using namespace std;#define ZERO 1e-6struct P{    double x, y;};double ans;/*求出a、b两点间的距离*/double dis(P a, P b){    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}/*对角线ab与对角线cd相交时，求出最短距离和*/void getcross(P a, P b, P c, P d){    double a1 = b.y - a.y, b1 = a.x - b.x, c1 = a.x * b.y - b.x * a.y;    double a2 = d.y - c.y, b2 = c.x - d.x, c2 = c.x * d.y - d.x * c.y;    if(fabs(a1 * b2 - a2 * b1) < ZERO) return ;    P p;    p.y = (a2 * c1 - a1 * c2) / (a2 * b1 - a1 * b2);    p.x = (b2 * c1 - b1 * c2) / (a1 * b2 - a2 * b1);    ans = min(ans, dis(a, p) + dis(b, p) + dis(c, p) + dis(d, p));}int main(){    P a, b, c, d;    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &a.x, &a.y,                 &b.x, &b.y, &c.x, &c.y, &d.x, &d.y))     {        if(fabs(a.x + 1) < ZERO) break;        ans = 1e8;        /*        这是凹四边形的情况，        凹点要么是a，要么是b，要么是c，要么是d        */        ans = min(ans, dis(b, a) + dis(c, a) + dis(d, a));        ans = min(ans, dis(a, b) + dis(c, b) + dis(d, b));        ans = min(ans, dis(a, c) + dis(b, c) + dis(d, c));        ans = min(ans, dis(a, d) + dis(b, d) + dis(c ,d));        /*        这是凸四边形的情况；        对角线要么是ab，要么ac，要么是ad        */        getcross(a, b, c, d);        getcross(a, d, b, c);        getcross(a, c, b, d);        printf("%.4f\n", ans);     }    return 0;}