poj1673 EXOCENTER OF A TRIANGLE

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地址:http://poj.org/problem?id=1673

题目:

EXOCENTER OF A TRIANGLE
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 3637   Accepted: 1467

Description

Given a triangle ABC, the Extriangles of ABC are constructed as follows: 
On each side of ABC, construct a square (ABDE, BCHJ and ACFG in the figure below). 
Connect adjacent square corners to form the three Extriangles (AGD, BEJ and CFH in the figure). 
The Exomedians of ABC are the medians of the Extriangles, which pass through vertices of the original triangle,extended into the original triangle (LAO, MBO and NCO in the figure. As the figure indicates, the three Exomedians intersect at a common point called the Exocenter (point O in the figure). 
This problem is to write a program to compute the Exocenters of triangles. 
技术分享

Input

The first line of the input consists of a positive integer n, which is the number of datasets that follow. Each dataset consists of 3 lines; each line contains two floating point values which represent the (two -dimensional) coordinate of one vertex of a triangle. So, there are total of (n*3) + 1 lines of input. Note: All input triangles wi ll be strongly non-degenerate in that no vertex will be within one unit of the line through the other two vertices.

Output

For each dataset you must print out the coordinates of the Exocenter of the input triangle correct to four decimal places.

Sample Input

2
0.0 0.0
9.0 12.0
14.0 0.0
3.0 4.0
13.0 19.0
2.0 -10.0

Sample Output

9.0000 3.7500
-48.0400 23.3600

Source

 
思路:
  其实就是重心。
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <algorithm>
  5 
  6 
  7 using namespace std;
  8 const double PI = acos(-1.0);
  9 const double eps = 1e-10;
 10 
 11 /****************常用函数***************/
 12 //判断ta与tb的大小关系
 13 int sgn( double ta, double tb)
 14 {
 15     if(fabs(ta-tb)<eps)return 0;
 16     if(ta<tb)   return -1;
 17     return 1;
 18 }
 19 
 20 //
 21 class Point
 22 {
 23 public:
 24 
 25     double x, y;
 26 
 27     Point(){}
 28     Point( double tx, double ty){ x = tx, y = ty;}
 29 
 30     bool operator < (const Point &_se) const
 31     {
 32         return x<_se.x || (x==_se.x && y<_se.y);
 33     }
 34     friend Point operator + (const Point &_st,const Point &_se)
 35     {
 36         return Point(_st.x + _se.x, _st.y + _se.y);
 37     }
 38     friend Point operator - (const Point &_st,const Point &_se)
 39     {
 40         return Point(_st.x - _se.x, _st.y - _se.y);
 41     }
 42     //点位置相同(double类型)
 43     bool operator == (const Point &_off)const
 44     {
 45         return  sgn(x, _off.x) == 0 && sgn(y, _off.y) == 0;
 46     }
 47 
 48 };
 49 
 50 /****************常用函数***************/
 51 //点乘
 52 double dot(const Point &po,const Point &ps,const Point &pe)
 53 {
 54     return (ps.x - po.x) * (pe.x - po.x) + (ps.y - po.y) * (pe.y - po.y);
 55 }
 56 //叉乘
 57 double xmult(const Point &po,const Point &ps,const Point &pe)
 58 {
 59     return (ps.x - po.x) * (pe.y - po.y) - (pe.x - po.x) * (ps.y - po.y);
 60 }
 61 //两点间距离的平方
 62 double getdis2(const Point &st,const Point &se)
 63 {
 64     return (st.x - se.x) * (st.x - se.x) + (st.y - se.y) * (st.y - se.y);
 65 }
 66 //两点间距离
 67 double getdis(const Point &st,const Point &se)
 68 {
 69     return sqrt((st.x - se.x) * (st.x - se.x) + (st.y - se.y) * (st.y - se.y));
 70 }
 71 
 72 //两点表示的向量
 73 class Line
 74 {
 75 public:
 76 
 77     Point s, e;//两点表示,起点[s],终点[e]
 78     double a, b, c;//一般式,ax+by+c=0
 79     double angle;//向量的角度,[-pi,pi]
 80 
 81     Line(){}
 82     Line( Point ts, Point te):s(ts),e(te){}//get_angle();}
 83     Line(double _a,double _b,double _c):a(_a),b(_b),c(_c){}
 84 
 85     //排序用
 86     bool operator < (const Line &ta)const
 87     {
 88         return angle<ta.angle;
 89     }
 90     //向量与向量的叉乘
 91     friend double operator / ( const Line &_st, const  Line &_se)
 92     {
 93         return (_st.e.x - _st.s.x) * (_se.e.y - _se.s.y) - (_st.e.y - _st.s.y) * (_se.e.x - _se.s.x);
 94     }
 95     //向量间的点乘
 96     friend double operator *( const Line &_st, const  Line &_se)
 97     {
 98         return (_st.e.x - _st.s.x) * (_se.e.x - _se.s.x) - (_st.e.y - _st.s.y) * (_se.e.y - _se.s.y);
 99     }
100     //从两点表示转换为一般表示
101     //a=y2-y1,b=x1-x2,c=x2*y1-x1*y2
102     bool pton()
103     {
104         a = e.y - s.y;
105         b = s.x - e.x;
106         c = e.x * s.y - e.y * s.x;
107         return true;
108     }
109     //半平面交用
110     //点在向量左边(右边的小于号改成大于号即可,在对应直线上则加上=号)
111     friend bool operator < (const Point &_Off, const Line &_Ori)
112     {
113         return (_Ori.e.y - _Ori.s.y) * (_Off.x - _Ori.s.x)
114             < (_Off.y - _Ori.s.y) * (_Ori.e.x - _Ori.s.x);
115     }
116     //求直线或向量的角度
117     double get_angle( bool isVector = true)
118     {
119         angle = atan2( e.y - s.y, e.x - s.x);
120         if(!isVector && angle < 0)
121             angle += PI;
122         return angle;
123     }
124 
125     //点在线段或直线上 1:点在直线上 2点在s,e所在矩形内
126     bool has(const Point &_Off, bool isSegment = false) const
127     {
128         bool ff = sgn( xmult( s, e, _Off), 0) == 0;
129         if( !isSegment) return ff;
130         return ff
131             && sgn(_Off.x - min(s.x, e.x), 0) >= 0 && sgn(_Off.x - max(s.x, e.x), 0) <= 0
132             && sgn(_Off.y - min(s.y, e.y), 0) >= 0 && sgn(_Off.y - max(s.y, e.y), 0) <= 0;
133     }
134 
135     //点到直线/线段的距离
136     double dis(const Point &_Off, bool isSegment = false)
137     {
138         ///化为一般式
139         pton();
140         //到直线垂足的距离
141         double td = (a * _Off.x + b * _Off.y + c) / sqrt(a * a + b * b);
142         //如果是线段判断垂足
143         if(isSegment)
144         {
145             double xp = (b * b * _Off.x - a * b * _Off.y - a * c) / ( a * a + b * b);
146             double yp = (-a * b * _Off.x + a * a * _Off.y - b * c) / (a * a + b * b);
147             double xb = max(s.x, e.x);
148             double yb = max(s.y, e.y);
149             double xs = s.x + e.x - xb;
150             double ys = s.y + e.y - yb;
151            if(xp > xb + eps || xp < xs - eps || yp > yb + eps || yp < ys - eps)
152                 td = min( getdis(_Off,s), getdis(_Off,e));
153         }
154         return fabs(td);
155     }
156 
157     //关于直线对称的点
158     Point mirror(const Point &_Off)
159     {
160         ///注意先转为一般式
161         Point ret;
162         double d = a * a + b * b;
163         ret.x = (b * b * _Off.x - a * a * _Off.x - 2 * a * b * _Off.y - 2 * a * c) / d;
164         ret.y = (a * a * _Off.y - b * b * _Off.y - 2 * a * b * _Off.x - 2 * b * c) / d;
165         return ret;
166     }
167     //计算两点的中垂线
168     static Line ppline(const Point &_a,const Point &_b)
169     {
170         Line ret;
171         ret.s.x = (_a.x + _b.x) / 2;
172         ret.s.y = (_a.y + _b.y) / 2;
173         //一般式
174         ret.a = _b.x - _a.x;
175         ret.b = _b.y - _a.y;
176         ret.c = (_a.y - _b.y) * ret.s.y + (_a.x - _b.x) * ret.s.x;
177         //两点式
178         if(fabs(ret.a) > eps)
179         {
180             ret.e.y = 0.0;
181             ret.e.x = - ret.c / ret.a;
182             if(ret.e == ret. s)
183             {
184                 ret.e.y = 1e10;
185                 ret.e.x = - (ret.c - ret.b * ret.e.y) / ret.a;
186             }
187         }
188         else
189         {
190             ret.e.x = 0.0;
191             ret.e.y = - ret.c / ret.b;
192             if(ret.e == ret. s)
193             {
194                 ret.e.x = 1e10;
195                 ret.e.y = - (ret.c - ret.a * ret.e.x) / ret.b;
196             }
197         }
198         return ret;
199     }
200 
201     //------------直线和直线(向量)-------------
202     //向量向左边平移t的距离
203     Line& moveLine( double t)
204     {
205         Point of;
206         of = Point( -( e.y - s.y), e.x - s.x);
207         double dis = sqrt( of.x * of.x + of.y * of.y);
208         of.x= of.x * t / dis, of.y = of.y * t / dis;
209         s = s + of, e = e + of;
210         return *this;
211     }
212     //直线重合
213     static bool equal(const Line &_st,const Line &_se)
214     {
215         return _st.has( _se.e) && _se.has( _st.s);
216     }
217     //直线平行
218     static bool parallel(const Line &_st,const Line &_se)
219     {
220         return sgn( _st / _se, 0) == 0;
221     }
222     //两直线(线段)交点
223     //返回-1代表平行,0代表重合,1代表相交
224     static bool crossLPt(const Line &_st,const Line &_se, Point &ret)
225     {
226         if(parallel(_st,_se))
227         {
228             if(Line::equal(_st,_se)) return 0;
229             return -1;
230         }
231         ret = _st.s;
232         double t = ( Line(_st.s,_se.s) / _se) / ( _st / _se);
233         ret.x += (_st.e.x - _st.s.x) * t;
234         ret.y += (_st.e.y - _st.s.y) * t;
235         return 1;
236     }
237     //------------线段和直线(向量)----------
238     //直线和线段相交
239     //参数:直线[_st],线段[_se]
240     friend bool crossSL( Line &_st, Line &_se)
241     {
242         return sgn( xmult( _st.s, _se.s, _st.e) * xmult( _st.s, _st.e, _se.e), 0) >= 0;
243     }
244 
245     //判断线段是否相交(注意添加eps)
246     static bool isCrossSS( const Line &_st, const  Line &_se)
247     {
248         //1.快速排斥试验判断以两条线段为对角线的两个矩形是否相交
249         //2.跨立试验(等于0时端点重合)
250         return
251             max(_st.s.x, _st.e.x) >= min(_se.s.x, _se.e.x) &&
252             max(_se.s.x, _se.e.x) >= min(_st.s.x, _st.e.x) &&
253             max(_st.s.y, _st.e.y) >= min(_se.s.y, _se.e.y) &&
254             max(_se.s.y, _se.e.y) >= min(_st.s.y, _st.e.y) &&
255             sgn( xmult( _se.s, _st.s, _se.e) * xmult( _se.s, _se.e, _st.s), 0) >= 0 &&
256             sgn( xmult( _st.s, _se.s, _st.e) * xmult( _st.s, _st.e, _se.s), 0) >= 0;
257     }
258 };
259 
260 //寻找凸包的graham 扫描法所需的排序函数
261 Point gsort;
262 bool gcmp( const Point &ta, const Point &tb)/// 选取与最后一条确定边夹角最小的点,即余弦值最大者
263 {
264     double tmp = xmult( gsort, ta, tb);
265     if( fabs( tmp) < eps)
266         return getdis( gsort, ta) < getdis( gsort, tb);
267     else if( tmp > 0)
268         return 1;
269     return 0;
270 }
271 
272 
273 
274 class triangle
275 {
276 public:
277     Point a, b, c;//顶点
278     triangle(){}
279     triangle(Point a, Point b, Point c): a(a), b(b), c(c){}
280 
281     //计算三角形面积
282     double area()
283     {
284         return fabs( xmult(a, b, c)) / 2.0;
285     }
286 
287     //计算三角形外心
288     //返回:外接圆圆心
289     Point circumcenter()
290     {
291         double pa = a.x * a.x + a.y * a.y;
292         double pb = b.x * b.x + b.y * b.y;
293         double pc = c.x * c.x + c.y * c.y;
294         double ta = pa * ( b.y - c.y) - pb * ( a.y - c.y) + pc * ( a.y - b.y);
295         double tb = -pa * ( b.x - c.x) + pb * ( a.x - c.x) - pc * ( a.x - b.x);
296         double tc = a.x * ( b.y - c.y) - b.x * ( a.y - c.y) + c.x * ( a.y - b.y);
297         return Point( ta / 2.0 / tc, tb / 2.0 / tc);
298     }
299 
300     //计算三角形内心
301     //返回:内接圆圆心
302     Point incenter()
303     {
304         Line u, v;
305         double m, n;
306         u.s = a;
307         m = atan2(b.y - a.y, b.x - a.x);
308         n = atan2(c.y - a.y, c.x - a.x);
309         u.e.x = u.s.x + cos((m + n) / 2);
310         u.e.y = u.s.y + sin((m + n) / 2);
311         v.s = b;
312         m = atan2(a.y - b.y, a.x - b.x);
313         n = atan2(c.y - b.y, c.x - b.x);
314         v.e.x = v.s.x + cos((m + n) / 2);
315         v.e.y = v.s.y + sin((m + n) / 2);
316         Point ret;
317         Line::crossLPt(u,v,ret);
318         return ret;
319     }
320 
321     //计算三角形垂心
322     //返回:高的交点
323     Point perpencenter()
324     {
325         Line u,v;
326         u.s = c;
327         u.e.x = u.s.x - a.y + b.y;
328         u.e.y = u.s.y + a.x - b.x;
329         v.s = b;
330         v.e.x = v.s.x - a.y + c.y;
331         v.e.y = v.s.y + a.x - c.x;
332         Point ret;
333         Line::crossLPt(u,v,ret);
334         return ret;
335     }
336 
337     //计算三角形重心
338     //返回:重心
339     //到三角形三顶点距离的平方和最小的点
340     //三角形内到三边距离之积最大的点
341     Point barycenter()
342     {
343         Line u,v;
344         u.s.x = (a.x + b.x) / 2;
345         u.s.y = (a.y + b.y) / 2;
346         u.e = c;
347         v.s.x = (a.x + c.x) / 2;
348         v.s.y = (a.y + c.y) / 2;
349         v.e = b;
350         Point ret;
351         Line::crossLPt(u,v,ret);
352         return ret;
353     }
354 
355     //计算三角形费马点
356     //返回:到三角形三顶点距离之和最小的点
357     Point fermentPoint()
358     {
359         Point u, v;
360         double step = fabs(a.x) + fabs(a.y) + fabs(b.x) + fabs(b.y) + fabs(c.x) + fabs(c.y);
361         int i, j, k;
362         u.x = (a.x + b.x + c.x) / 3;
363         u.y = (a.y + b.y + c.y) / 3;
364         while (step > eps)
365         {
366             for (k = 0; k < 10; step /= 2, k ++)
367             {
368                 for (i = -1; i <= 1; i ++)
369                 {
370                     for (j =- 1; j <= 1; j ++)
371                     {
372                         v.x = u.x + step * i;
373                         v.y = u.y + step * j;
374                         if (getdis(u,a) + getdis(u,b) + getdis(u,c) > getdis(v,a) + getdis(v,b) + getdis(v,c))
375                             u = v;
376                     }
377                 }
378             }
379         }
380         return u;
381     }
382 };
383 
384 triangle tr;
385 int main(void)
386 {
387     int n;
388     scanf("%d",&n);
389     while(n--)
390     {
391         scanf("%lf%lf%lf%lf%lf%lf",&tr.a.x,&tr.a.y,&tr.b.x,&tr.b.y,&tr.c.x,&tr.c.y);
392         Point ans=tr.perpencenter();
393         printf("%.4f %.4f\n",ans.x,ans.y);
394     }
395     return 0;
396 }

 

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