poj1673 EXOCENTER OF A TRIANGLE
Posted weeping
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了poj1673 EXOCENTER OF A TRIANGLE相关的知识,希望对你有一定的参考价值。
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.
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.
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 //点 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 }
以上是关于poj1673 EXOCENTER OF A TRIANGLE的主要内容,如果未能解决你的问题,请参考以下文章
POJ 1286Necklace of Beads(polya定理)
poj 1958 Strange Towers of Hanoi
BZOJ 1673 [Usaco2005 Dec]Scales 天平:dfs 启发式搜索 A*搜索