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

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。

给出三角形三点坐标，求外接圆圆心和半径。
给出三角形三点坐标，求内切圆圆心和半径。
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°)
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。
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。
给出两个相离的圆，求半径为r且与两圆都相切的圆。

分析：

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。
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。
分点在圆内，圆上，圆外三种情况，注意最终结果的范围。
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。
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。
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;    //åœ†å¿ƒ
189     double r;    //å?Šå¾„
190     Point point(double a)
191     {
192         return Point(c.x+r*cos(a), c.y+r*sin(a));
193     }
194 };
195 //ä¸¤åœ†ç›¸äº¤å¹¶è¿”å›?äº¤ç‚¹ä¸ªæ•°
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;        //åˆ¤åˆ«å¼?
202     if(dcmp(delta) < 0)    return 0;    //ç›¸ç¦»
203     if(dcmp(delta) == 0)            //ç›¸åˆ‡
204     {
205         t1 = t2 = -f / (2 * e);
206         sol.push_back(L.point(t1));
207         return 1;
208     }
209     //ç›¸äº¤
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

时间： 2024-10-08 19:08:20

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