UVa 12304 (6个二维几何问题合集) 2D Geometry 110 in 1!

这个题能1A纯属运气，要是WA掉，可真不知道该怎么去调了。

题意：

这是完全独立的6个子问题。代码中是根据字符串的长度来区分问题编号的。

1. 给出三角形三点坐标，求外接圆圆心和半径。
2. 给出三角形三点坐标，求内切圆圆心和半径。
3. 给出一个圆和一个定点，求过定点作圆的所有切线的倾角(0≤a＜180°)
4. 给出一个点和一条直线，求一个半径为r的过该点且与该直线相切的圆。
5. 给出两条相交直线，求所有半径为r且与两直线都相切的圆。
6. 给出两个相离的圆，求半径为r且与两圆都相切的圆。

分析：

1. 写出三角形两边的垂直平分线的一般方程(注意去掉分母，避免直线是水平或垂直的特殊情况)，然后联立求解即可。
2. 有一个很简洁的三角形内心坐标公式(证明有点复杂，可用向量来证，其中多次用到角平分线定理)，公式详见代码。
3. 分点在圆内，圆上，圆外三种情况，注意最终结果的范围。
4. 到定点距离为r的轨迹是个圆，与直线相切的圆心的轨迹是两条平行直线。最终转化为求圆与两条平行线的交点。
5. 我开始用的方法是求出圆心到两直线交点的距离，以及与其中一条直线的夹角，依次旋转三个90°即可得到另外三个点。但是对比正确结果，误差居然达到了个位(如果代码没有错的话)！后来参考了lrj的思路，就是讲两直线分别向两侧平移r距离，这样得到的四条直线两两相交得到的四个交点就是所求。
6. 看起来有点复杂，仔细分析，半径为r与圆外切的圆心的轨迹还是个圆。因此问题转化为求半径扩大以后的两圆的交点。

体会：

• (Point)(x, y)是强制类型转换，Point(x, y)才是调用构造函数。前者只会将x的值复制，y的值则是默认值0.
• 计算的中间步骤越多，误差越大，最好能优化算法，或者调整EPS的大小。

 

  1 //#define LOCAL

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <string>

  5 #include <algorithm>

  6 #include <cmath>

  7 #include <vector>

  8 using namespace std;

  9 

 10 struct Point

 11 {

 12     double x, y;

 13     Point(double xx=0, double yy=0) :x(xx),y(yy) {}

 14 };

 15 typedef Point Vector;

 16 

 17 Point read_point(void)

 18 {

 19     double x, y;

 20     scanf("%lf%lf", &x, &y);

 21     return Point(x, y);

 22 }

 23 

 24 const double EPS = 1e-7;

 25 const double PI = acos(-1.0);

 26 

 27 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }

 28 

 29 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); }

 30 

 31 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); }

 32 

 33 Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); }

 34 

 35 bool operator < (const Point& a, const Point& b)

 36 { return a.x < b.x || (a.x == b.x && a.y < b.y); }

 37 

 38 int dcmp(double x)

 39 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; }

 40 

 41 bool operator == (const Point& a, const Point& b)

 42 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }

 43 

 44 double Dot(Vector A, Vector B)

 45 { return A.x*B.x + A.y*B.y; }

 46 

 47 double Length(Vector A)    { return sqrt(Dot(A, A)); }

 48 

 49 double Angle(Vector A, Vector B)

 50 { return acos(Dot(A, B) / Length(A) / Length(B)); }

 51 

 52 double Angle2(Vector A)    { return atan2(A.y, A.x); }

 53 

 54 Vector VRotate(Vector A, double rad)

 55 {

 56     return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));

 57 }

 58 

 59 Vector Normal(Vector A)

 60 {

 61     double l = Length(A);

 62     return Vector(-A.y/l, A.x/l);

 63 }

 64 

 65 double Change(double r)    { return r / PI * 180.0; }

 66 

 67 double Cross(Vector A, Vector B)

 68 { return A.x*B.y - A.y*B.x; }

 69 

 70 struct Circle

 71 {

 72     double x, y, r;

 73     Circle(double x=0, double y=0, double r=0):x(x), y(y), r(r) {}

 74     Point point(double a)

 75     {

 76         return Point(x+r*cos(a), y+r*sin(a));

 77     }

 78 };

 79 

 80 const int maxn = 1010;

 81 char s[maxn];

 82 

 83 int ID(char* s)

 84 {

 85     int l = strlen(s);

 86     switch(l)

 87     {

 88         case 19: return 0;

 89         case 15: return 1;

 90         case 23: return 2;

 91         case 46: return 3;

 92         case 33: return 4;

 93         case 43: return 5;

 94         default: return -1;

 95     }

 96 }

 97 

 98 void Solve(double A1, double B1, double C1, double A2, double B2, double C2, double& ansx, double& ansy)

 99 {

100     ansx = (B1*C2 - B2*C1) / (A1*B2 - A2*B1);

101     ansy = (C2*A1 - C1*A2) / (B1*A2 - B2*A1);    

102 }

103 

104 void problem0()

105 {

106     Point A, B, C;

107     scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y);

108     double A1 = B.x-A.x, B1 = B.y-A.y, C1 = (A.x*A.x-B.x*B.x+A.y*A.y-B.y*B.y)/2;

109     double A2 = C.x-A.x, B2 = C.y-A.y, C2 = (A.x*A.x-C.x*C.x+A.y*A.y-C.y*C.y)/2;

110     Point ans;

111     Solve(A1, B1, C1, A2, B2, C2, ans.x, ans.y);

112     double r = Length(ans - A);

113     printf("(%.6lf,%.6lf,%.6lf)\n", ans.x, ans.y, r);

114 }

115 

116 void problem1()

117 {

118     Point A, B, C;

119     scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y);

120     double a = Length(B-C), b = Length(A-C), c = Length(A-B);

121     double l = a+b+c;

122     Point ans = (A*a+B*b+C*c)/l;

123     double r = fabs(Cross(A-B, C-B)) / l;

124     printf("(%.6lf,%.6lf,%.6lf)\n", ans.x, ans.y, r);

125 }

126 

127 void problem2()

128 {

129     Circle C;

130     Point P, O;

131     scanf("%lf%lf%lf%lf%lf", &C.x, &C.y, &C.r, &P.x, &P.y);

132     double ans[2];

133     O.x = C.x, O.y = C.y;

134     double d = Length(P-O);

135     int k = dcmp(d-C.r);

136     if(k < 0)

137     {

138         printf("[]\n");

139         return;

140     }

141     else if(k == 0)

142     {

143         ans[0] = Change(Angle2(P-O)) + 90.0;

144         while(ans[0] >= 180.0)    ans[0] -= 180.0;

145         while(ans[0] < 0)        ans[0] += 180.0;

146         printf("[%.6lf]\n", ans[0]);

147         return;

148     }

149     else

150     {

151         double ag = asin(C.r/d);

152         double base = Angle2(P-O);

153         ans[0] = base + ag, ans[1] = base - ag;

154         ans[0] = Change(ans[0]), ans[1] = Change(ans[1]);

155         while(ans[0] >= 180.0)    ans[0] -= 180.0;

156         while(ans[0] < 0)        ans[0] += 180.0;

157         while(ans[1] >= 180.0)    ans[1] -= 180.0;

158         while(ans[1] < 0)        ans[1] += 180.0;

159         if(ans[0] >= ans[1])    swap(ans[0], ans[1]);

160         printf("[%.6lf,%.6lf]\n", ans[0], ans[1]);

161     }

162 }

163 

164 vector<Point> sol;

165 struct Line

166 {

167     Point p;

168     Vector v;

169     Line()    { }

170     Line(Point p, Vector v): p(p), v(v)    {}

171     Point point(double t)

172     {

173         return p + v*t;

174     }

175     Line move(double d)

176     {

177         return Line(p + Normal(v)*d, v);

178     }

179 };

180 Point GetIntersection(Line a, Line b)

181 {

182     Vector u = a.p - b.p;

183     double t = Cross(b.v, u) / Cross(a.v, b.v);

184     return a.p + a.v*t;

185 }

186 struct Circle2

187 {

188     Point c;    //圆心

189     double r;    //半径

190     Point point(double a)

191     {

192         return Point(c.x+r*cos(a), c.y+r*sin(a));

193     }

194 };

195 //两圆相交并返回交点个数 

196 int getLineCircleIntersection(Line L, Circle2 C, vector<Point>& sol)

197 {

198     double t1, t2;

199     double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;

200     double e = a*a + c*c, f = 2*(a*b + c*d), g = b*b + d*d - C.r*C.r;

201     double delta = f*f - 4*e*g;        //判别式

202     if(dcmp(delta) < 0)    return 0;    //相离

203     if(dcmp(delta) == 0)            //相切

204     {

205         t1 = t2 = -f / (2 * e);

206         sol.push_back(L.point(t1));

207         return 1;

208     }

209     //相交

210     t1 = (-f - sqrt(delta)) / (2 * e);    sol.push_back(L.point(t1));

211     t2 = (-f + sqrt(delta)) / (2 * e);    sol.push_back(L.point(t2));

212     return 2;

213 }

214 void problem3()

215 {

216     Circle2 C;

217     Point A, B;

218     scanf("%lf%lf%lf%lf%lf%lf%lf", &C.c.x, &C.c.y, &A.x, &A.y, &B.x, &B.y, &C.r);

219     Vector v = (A-B)/Length(A-B)*C.r;

220     //printf("%lf\n", Length(v));

221     Point p1 = A + Point(-v.y, v.x);

222     Point p2 = A + Point(v.y, -v.x);

223     //printf("%lf\n%lf", Length(p1-C.c), Length(p2-C.c));

224     Line L1(p1, v), L2(p2, v);

225     

226     sol.clear();

227     int cnt =  getLineCircleIntersection(L1, C, sol);

228     cnt += getLineCircleIntersection(L2, C, sol);

229     sort(sol.begin(), sol.end());

230     if(cnt == 0)    { printf("[]\n"); return; }

231     printf("[");

232     for(int i = 0; i < cnt-1; ++i)    printf("(%.6lf,%.6lf),", sol[i].x, sol[i].y);

233     printf("(%.6lf,%.6lf)]\n", sol[cnt-1].x, sol[cnt-1].y);

234 }

235 

236 void problem4()

237 {

238     double r;

239     Point A, B, C, D, E, ans[4];

240     scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y, &D.x, &D.y, &r);

241     Line a(A, B-A), b(C, D-C);

242     Line L1 = a.move(r), L2 = a.move(-r);

243     Line L3 = b.move(r), L4 = b.move(-r);

244     ans[0] = GetIntersection(L1, L3);

245     ans[1] = GetIntersection(L1, L4);

246     ans[2] = GetIntersection(L2, L3);

247     ans[3] = GetIntersection(L2, L4);

248     sort(ans, ans+4);

249     printf("[");

250     for(int i = 0; i < 3; ++i)    printf("(%.6lf,%.6lf),", ans[i].x, ans[i].y);

251     printf("(%.6lf,%.6lf)]\n", ans[3].x, ans[3].y);

252 }

253 

254 int getCircleCircleIntersection(Circle2 C1, Circle2 C2, vector<Point>& sol)

255 {

256     double d = Length(C1.c - C2.c);

257     if(dcmp(d) == 0)

258     {

259         if(dcmp(C1.r - C2.r) == 0)    return -1;

260         return 0;

261     }

262     if(dcmp(C1.r + C2.r - d) < 0)    return 0;

263     if(dcmp(fabs(C1.r - C2.r) - d) > 0)    return 0;

264 

265     double a = Angle2(C2.c - C1.c);

266     double da = acos((C1.r*C1.r + d*d - C2.r*C2.r) / (2*C1.r*d));

267     Point p1 = C1.point(a+da), p2 = C1.point(a-da);

268     sol.push_back(p1);

269     if(p1 == p2)    return 1;

270     sol.push_back(p2);

271     return 2;

272 }

273 

274 void problem5()

275 {

276     Circle2 C1, C2;

277     double r;

278     vector<Point> sol;

279     scanf("%lf%lf%lf%lf%lf%lf%lf", &C1.c.x, &C1.c.y, &C1.r, &C2.c.x, &C2.c.y, &C2.r, &r);

280     double d = Length(C1.c - C2.c);

281     C1.r += r, C2.r += r;

282     if(dcmp(C1.r+C2.r-d) < 0)    { printf("[]\n"); return; }

283     int n = getCircleCircleIntersection(C1, C2, sol);

284     sort(sol.begin(), sol.end());

285     printf("[");

286     for(int i = 0; i < n-1; ++i)    printf("(%.6lf,%.6lf),", sol[i].x, sol[i].y);

287     printf("(%.6lf,%.6lf)]\n", sol[n-1].x, sol[n-1].y);

288 }

289 

290 int main()

291 {

292     #ifdef    LOCAL

293         freopen("12304in.txt", "r", stdin);

294     #endif

295 

296     while(scanf("%s", s) == 1)

297     {

298         int proID = ID(s);

299         switch(proID)

300         {

301             case 0:    problem0();    break;

302             case 1:    problem1();    break;

303             case 2:    problem2();    break;

304             case 3:    problem3();    break;

305             case 4:    problem4();    break;

306             case 5:    problem5();    break;

307             default: break;

308         }

309     }

310     

311     return 0;

312 }

代码君