HDU 3694 Fermat Point in Quadrangle(四边形的费马点)

Fermat Point in Quadrangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1460    Accepted Submission(s): 241

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

 

Source

2010 Asia Fuzhou Regional Contest

 

Recommend

axubiao

 

这个题目wa的我有点伤心，偶尔犯一些小错误，在别的类型题目可能是容易发现的，但是在计算几何里面不但不容易发现而且是致命的

拿到这个题目想都没想模拟退火的思路，很快写出代码（下下面会有） 老是过不了，不知道为什么四变形就不可以！

然后就想这样子做

首先按照点排序，和凸包一样的排序，判断四点是否共线，不成立后判断是否是凸包！

我的解决方案有点极端：

首先四个点枚举，每一个点作为费马点一次求出距离，这个是假设不是凸包的情况下

然后在求是凸包的情况下，防止错误两者取最小的保证正确

也就是说费马点不是对角线的交点就是四个点中的某一个，尝试一下取最下

至于下面代码为什么这么长，因为老是wa老是改，就成这样了！

#include <iostream>#include <string.h>#include <stdio.h>#include <math.h>#include <algorithm>using namespace std;struct point{double x;double y;}po[100],my_p;struct line{point a;point b;};double dis(point &a,point &b){return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));}double across(point &a,point &b,point &c){return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}int cmp(const void *a,const void *b){return across(po[0],*(point*)a,*(point*)b) > 1e-8 ? -1 : 1;}int is_convex(int num,point *p){int i;p[num]=p[0];num++;p[num]=po[1];num++;int count=0;for(i=2;i < num;i++){if(across(p[i-2],p[i-1],p[i]) < 1e-8){my_p=p[i-1];return 0;}}return 1;}point intersection(line &u,line &v){point ret=u.a;double t=((u.a.x-v.a.x)*(v.a.y-v.b.y) - (u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));ret.x+=(u.b.x-u.a.x)*t;ret.y+=(u.b.y-u.a.y)*t;return ret;}bool zero(double a)//判断结果是否为0{return fabs(a) <= 1e-8;}bool parallel(point &u1,point &u2,point &v1,point &v2)//判断直线u1 u2 and v1 v2 是否平行{return zero((u1.x-u2.x)*(v1.y-v2.y)-(v1.x-v2.x)*(u1.y-u2.y));}bool equal(point &a,point &b){return fabs(a.x-b.x) < 1e-8 && fabs(a.y-b.y) < 1e-8;}double MIN(double a,double b){return a < b ? a : b;}int main(){int i,j,k;point temp;point my_temp;line a,b;double ans,ans1,tt;int pos;while(1){scanf("%lf%lf",&po[0].x,&po[0].y);temp=po[0];pos=0;for(i=1;i<4;i++){scanf("%lf%lf",&po[i].x,&po[i].y);if(temp.y > po[i].y)temp=po[i],pos=i;}if(po[0].x==-1 && po[0].y==-1 && po[1].x==-1 && po[1].y==-1 && po[2].x==-1 && po[2].y==-1 && po[3].x==-1 && po[3].y==-1)return 0;my_temp=po[0];po[0]=po[pos];po[pos]=my_temp;qsort(po+1,3,sizeof(po[0]),cmp);ans=ans1=100000000;tt=0;for(i=0;i<4;i++){tt=0;for(j=0;j<4;j++)tt+=dis(po[i],po[j]);ans1=MIN(tt,ans1);}if(parallel(po[0],po[1],po[1],po[2]) && parallel(po[1],po[2],po[2],po[3]))NULL;else if(is_convex(4,po)){a.a=po[0],a.b=po[2];b.a=po[1],b.b=po[3];ans=0;my_p=intersection(a,b);for(i=0;i<4;i++)ans+=dis(my_p,po[i]);//printf("%lf %lf \n",my_p.x,my_p.y);}printf("%.4lf\n",MIN(ans,ans1));}return 0;}

模拟退火代码：

#include <iostream>#include <stdio.h>#include <cmath>#include <string.h>using namespace std;#define inf 1e-10struct point{double x;double y;}po[4];double dis(point &a,point &b){return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);}double find_ans(){int i;point temp,p,q;double ans=0;temp=p=q=po[0];for(i=1;i<4;i++)ans+=dis(temp,po[i]);double step=1500,rec;while(step > inf){q=p;temp=q;rec=0;temp.x+=step;for(i=0;i<4;i++)rec+=dis(temp,po[i]);if(rec < ans){p=temp;ans=rec;}temp=q;rec=0;temp.x-=step;for(i=0;i<4;i++)rec+=dis(temp,po[i]);if(rec < ans){p=temp;ans=rec;}temp=q;rec=0;temp.y+=step;for(i=0;i<4;i++)rec+=dis(temp,po[i]);if(rec < ans){p=temp;ans=rec;}temp=q;rec=0;temp.y-=step;for(i=0;i<4;i++)rec+=dis(temp,po[i]);if(rec < ans){p=temp;ans=rec;}step/=2;}return ans;}int main(){int i,j,k;while(1){for(i=0;i<4;i++)scanf("%lf%lf",&po[i].x,&po[i].y);if(po[0].x==-1 && po[0].y==-1 && po[1].x==-1 && po[1].y==-1 && po[2].x==-1 && po[2].y==-1 && po[3].x==-1 && po[3].y==-1)return 0;printf("%.4lf\n",find_ans());}return 0;