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题意:要求解答6个关于圆的问题。
1.给出三角形坐标求外接圆
2.给出三角形坐标求内切圆
3.给出一个圆心和半径已知的圆,求过点(x,y)的所有和这个圆相切的直线
4.求所有和已知直线相切的过定点(x,y)的已知半径的圆的圆心
5.给出两个不平行的直线,求所有半径为r的同时和这两个直线相切的圆
6.给定两个相离的圆,求出所有和这两个圆外切的半径为r的圆。
比较恶心的计算几何模板题,直接模板吧。。。。
1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 #include<algorithm> 5 #include<iostream> 6 #include<memory.h> 7 #include<cstdlib> 8 #include<vector> 9 #define clc(a,b) memset(a,b,sizeof(a)) 10 #define LL long long int 11 #define up(i,x,y) for(i=x;i<=y;i++) 12 #define w(a) while(a) 13 using namespace std; 14 const int inf=0x3f3f3f3f; 15 const int N = 110; 16 const int maxn = 50; 17 const double eps = 1e-10; //调到1e-6以上第4问就可以用delta判断切线,但《训练指南》建议,尽量不要调eps 18 const double pi = acos(-1); 19 20 char type[maxn]; 21 22 int dcmp(double x) 23 { 24 return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1); 25 } 26 27 struct Point 28 { 29 double x; 30 double y; 31 32 Point(double x = 0, double y = 0):x(x), y(y) {} 33 34 bool operator < (const Point& e) const 35 { 36 return dcmp(x - e.x) < 0 || (dcmp(x - e.x) == 0 && dcmp(y - e.y) < 0); 37 } 38 39 bool operator == (const Point& e) const 40 { 41 return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0; 42 } 43 44 int read() 45 { 46 return scanf("%lf%lf", &x, &y); 47 } 48 } p[3]; 49 50 typedef Point Vector; 51 52 Vector operator + (Point A, Point B) 53 { 54 return Vector(A.x + B.x, A.y + B.y); 55 } 56 57 Vector operator - (Point A, Point B) 58 { 59 return Vector(A.x - B.x, A.y - B.y); 60 } 61 62 Vector operator * (Point A, double p) 63 { 64 return Vector(A.x * p, A.y * p); 65 } 66 67 Vector operator / (Point A, double p) 68 { 69 return Vector(A.x / p, A.y / p); 70 } 71 72 struct Line 73 { 74 Point p; 75 Point v; 76 77 Line() {} 78 Line(Point p, Point v):p(p), v(v) {} 79 80 int read() 81 { 82 return scanf("%lf%lf%lf%lf", &p.x, &p.y, &v.x, &v.y); 83 } 84 85 Point point(double t) 86 { 87 return p + v * t; 88 } 89 }; 90 91 struct Circle 92 { 93 Point c; 94 double r; 95 96 Circle() {} 97 Circle(Point c, double r):c(c), r(r) {} 98 99 int read() 100 { 101 return scanf("%lf%lf%lf", &c.x, &c.y, &r); 102 } 103 104 Point point(double a) 105 { 106 return Point(c.x + r * cos(a), c.y + r * sin(a)); 107 } 108 }; 109 110 double Dot(Vector A, Vector B) 111 { 112 return A.x * B.x + A.y * B.y; 113 } 114 115 double Cross(Vector A, Vector B) 116 { 117 return A.x * B.y - B.x * A.y; 118 } 119 120 double Length(Vector A) 121 { 122 return sqrt(Dot(A, A)); 123 } 124 125 Vector Rotate(Vector A, double rad) 126 { 127 return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); 128 } 129 130 Vector Normal(Vector A) 131 { 132 double L = Length(A); 133 return Vector(-A.y / L, A.x / L); 134 } 135 136 double DistanceToLine(Point P, Point A, Point B) //点到直线的距离 137 { 138 Vector v1 = B - A; 139 Vector v2 = P - A; 140 return fabs(Cross(v1, v2) / Length(v1)); 141 } 142 143 double angle(Vector v) //求向量的极角 144 { 145 return atan2(v.y, v.x); 146 } 147 148 Point GetLineIntersection(Line l1, Line l2) //求两直线的交点(前提:相交) 149 { 150 Vector u = l1.p - l2.p; 151 double t = Cross(l2.v, u) / Cross(l1.v, l2.v); 152 return l1.point(t); 153 } 154 155 int getLineCircleIntersection(Line l, Circle C, double& t1, double& t2, vector<Point>& sol) //求直线与圆的交点 156 { 157 double a = l.v.x; 158 double b = l.p.x - C.c.x; 159 double c = l.v.y; 160 double d = l.p.y - C.c.y; 161 double e = a * a + c * c; 162 double f = 2 * (a * b + c * d); 163 double g = b * b + d * d - C.r * C.r; 164 double delta = f * f - 4 * e * g; 165 double dist = DistanceToLine(C.c, l.p, l.p+l.v); 166 if(dcmp(dist - C.r) == 0) //相切,此处需特殊判断,不能用delta 167 { 168 t1 = t2 = -f / (2 * e); 169 sol.push_back(l.point(t1)); 170 return 1; 171 } 172 if(dcmp(delta) < 0) return 0; //相离 173 else //相交 174 { 175 t1 = (-f - sqrt(delta)) / (2 * e); 176 sol.push_back(l.point(t1)); 177 t2 = (-f + sqrt(delta)) / (2 * e); 178 sol.push_back(l.point(t2)); 179 return 2; 180 } 181 } 182 183 int GetCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol) //求圆与圆的交点 184 { 185 double d = Length(C1.c - C2.c); 186 if(dcmp(d) == 0) 187 { 188 if(dcmp(C1.r - C2.r) == 0) return -1; //两圆重合 189 return 0; //同心圆但不重合 190 } 191 if(dcmp(C1.r + C2.r - d) < 0) return 0; //外离 192 if(dcmp(fabs(C1.r - C2.r) - d) > 0) return 0; //内含 193 double a = angle(C2.c - C1.c); 194 double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d)); 195 Point p1 = C1.point(a + da); 196 Point p2 = C1.point(a - da); 197 sol.push_back(p1); 198 if(p1 == p2) return 1; //外切 199 sol.push_back(p2); 200 return 2; 201 } 202 203 Circle CircumscribedCircle(Point p1, Point p2, Point p3) //求三角形的外心 204 { 205 double Bx = p2.x - p1.x, By = p2.y - p1.y; 206 double Cx = p3.x - p1.x, Cy = p3.y - p1.y; 207 double D = 2 * (Bx * Cy - By * Cx); 208 double cx = (Cy * (Bx * Bx + By * By) - By * (Cx * Cx + Cy * Cy)) / D + p1.x; 209 double cy = (Bx * (Cx * Cx + Cy * Cy) - Cx * (Bx * Bx + By * By)) / D + p1.y; 210 Point p(cx, cy); 211 return Circle(p, Length(p1-p)); 212 } 213 214 Circle InscribedCircle(Point p1, Point p2, Point p3) //求三角形的内切圆 215 { 216 double a = Length(p3 - p2); 217 double b = Length(p3 - p1); 218 double c = Length(p2 - p1); 219 Point p = (p1 * a + p2 * b + p3 * c) / (a + b + c); 220 return Circle(p, DistanceToLine(p, p2, p3)); 221 } 222 223 int TangentLineThroughPoint(Point p, Circle C, Vector *v) //求点到圆的直线 224 { 225 Vector u = C.c - p; 226 double dist = Length(u); 227 if(dcmp(dist - C.r) < 0) return 0; 228 else if(dcmp(dist - C.r) < eps) 229 { 230 v[0] = Rotate(u, pi / 2); 231 return 1; 232 } 233 else 234 { 235 double ang = asin(C.r / dist); 236 v[0] = Rotate(u, ang); 237 v[1] = Rotate(u, -ang); 238 return 2; 239 } 240 } 241 242 void CircleThroughAPointAndTangentToALineWithRadius(Point p, Point p1, Point p2, double r) 243 { 244 Vector AB = p2 - p1; 245 Vector change1 = Rotate(AB, pi / 2) / Length(AB) * r; 246 Vector change2 = Rotate(AB, -pi / 2) / Length(AB) * r; 247 Line l1(p1 + change1, AB); 248 Line l2(p1 + change2, AB); 249 vector<Point> sol; 250 sol.clear(); 251 double t1, t2; 252 int cnt1 = getLineCircleIntersection(l1, Circle(p, r), t1, t2, sol); 253 int cnt2 = getLineCircleIntersection(l2, Circle(p, r), t1, t2, sol); 254 int cnt = cnt1 + cnt2; 255 if(cnt) sort(sol.begin(), sol.end()); 256 printf("["); 257 for(int i = 0; i < cnt; i++) 258 { 259 printf("(%.6f,%.6f)", sol[i].x, sol[i].y); 260 if(cnt == 2 && !i) printf(","); 261 } 262 puts("]"); 263 } 264 265 void CircleTangentToTwoLinesWithRadius(Point A, Point B, Point C, Point D, double r) 266 { 267 Vector AB = B - A; 268 Vector change = Normal(AB) * r; 269 Point newA1 = A + change; 270 Point newA2 = A - change; 271 Vector CD = D - C; 272 Vector update = Normal(CD) * r; 273 Point newC1 = C + update; 274 Point newC2 = C - update; 275 Point p[5]; 276 p[0] = GetLineIntersection(Line(newA1, AB), Line(newC1, CD)); 277 p[1] = GetLineIntersection(Line(newA1, AB), Line(newC2, CD)); 278 p[2] = GetLineIntersection(Line(newA2, AB), Line(newC1, CD)); 279 p[3] = GetLineIntersection(Line(newA2, AB), Line(newC2, CD)); 280 sort(p, p + 4); 281 printf("["); 282 printf("(%.6f,%.6f)", p[0].x, p[0].y); 283 for(int i = 1; i < 4; i++) 284 { 285 printf(",(%.6f,%.6f)", p[i].x, p[i].y); 286 } 287 puts("]"); 288 } 289 290 void CircleTangentToTwoDisjointCirclesWithRadius(Circle C1, Circle C2, double r) 291 { 292 Vector CC = C2.c - C1.c; 293 double rdist = Length(CC); 294 if(dcmp(2 * r - rdist + C1.r + C2.r) < 0) puts("[]"); 295 else if(dcmp(2 * r - rdist + C1.r + C2.r) == 0) 296 { 297 double ang = angle(CC); 298 Point A = C1.point(ang); 299 Point B = C2.point(ang + pi); 300 Point ret = (A + B) / 2; 301 printf("[(%.6f,%.6f)]\n", ret.x, ret.y); 302 } 303 else 304 { 305 Circle A = Circle(C1.c, C1.r + r); 306 Circle B = Circle(C2.c, C2.r + r); 307 vector<Point> sol; 308 sol.clear(); 309 GetCircleCircleIntersection(A, B, sol); 310 sort(sol.begin(), sol.end()); 311 printf("[(%.6f,%.6f),(%.6f,%.6f)]\n", sol[0].x, sol[0].y, sol[1].x, sol[1].y); 312 } 313 } 314 315 int main() 316 { 317 while(scanf("%s", type) == 1) 318 { 319 if(strcmp(type, "CircumscribedCircle") == 0) 320 { 321 Point p1, p2, p3; 322 p1.read(); 323 p2.read(); 324 p3.read(); 325 Circle ret = CircumscribedCircle(p1, p2, p3); 326 printf("(%f,%f,%f)\n", ret.c.x, ret.c.y, ret.r); 327 } 328 else if(strcmp(type, "InscribedCircle") == 0) 329 { 330 Point p1, p2, p3; 331 p1.read(); 332 p2.read(); 333 p3.read(); 334 Circle ret = InscribedCircle(p1, p2, p3); 335 printf("(%f,%f,%f)\n", ret.c.x, ret.c.y, ret.r); 336 } 337 else if(strcmp(type, "TangentLineThroughPoint") == 0) 338 { 339 Circle C; 340 Point p; 341 C.read(); 342 p.read(); 343 Vector v[3]; 344 int cnt = TangentLineThroughPoint(p, C, v); 345 double ret[3]; 346 for(int i = 0; i < cnt; i++) 347 { 348 ret[i] = angle(v[i]); 349 if(dcmp(ret[i] - pi) == 0) ret[i] = 0; 350 if(dcmp(ret[i]) < 0) ret[i] += pi; 351 } 352 sort(ret, ret + cnt); 353 printf("["); 354 for(int i = 0; i < cnt; i++) 355 { 356 printf("%.6f", ret[i] / pi * 180); 357 if(cnt == 2 && !i) printf(","); 358 } 359 puts("]"); 360 } 361 else if(strcmp(type, "CircleThroughAPointAndTangentToALineWithRadius") == 0) 362 { 363 Point p, p1, p2; 364 double r; 365 p.read(); 366 p1.read(); 367 p2.read(); 368 scanf("%lf", &r); 369 CircleThroughAPointAndTangentToALineWithRadius(p, p1, p2, r); 370 } 371 else if(strcmp(type, "CircleTangentToTwoLinesWithRadius") == 0) 372 { 373 Point A, B, C, D; 374 double r; 375 A.read(); 376 B.read(); 377 C.read(); 378 D.read(); 379 scanf("%lf", &r); 380 CircleTangentToTwoLinesWithRadius(A, B, C, D, r); 381 } 382 else 383 { 384 Circle C1, C2; 385 double r; 386 C1.read(); 387 C2.read(); 388 scanf("%lf", &r); 389 CircleTangentToTwoDisjointCirclesWithRadius(C1, C2, r); 390 } 391 } 392 return 0; 393 }
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原文地址:http://www.cnblogs.com/ITUPC/p/4870049.html