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。

 //#define LOCAL
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 using namespace std;

 struct Point
 {
     double x, y;
     Point(double xx=, double yy=) :x(xx),y(yy) {}
 };
 typedef Point Vector;

 Point read_point(void)
 {
     double x, y;
     scanf("%lf%lf", &x, &y);
     return Point(x, y);
 }

 const double EPS = 1e-;
 const double PI = acos(-1.0);

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

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

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

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

 bool operator < (const Point& a, const Point& b)
 { return a.x < b.x || (a.x == b.x && a.y < b.y); }

 int dcmp(double x)
 { if(fabs(x) < EPS) return ; else return x <  ? - : ; }

 bool operator == (const Point& a, const Point& b)
 { return dcmp(a.x-b.x) ==  && dcmp(a.y-b.y) == ; }

 double Dot(Vector A, Vector B)
 { return A.x*B.x + A.y*B.y; }

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

 double Angle(Vector A, Vector B)
 { return acos(Dot(A, B) / Length(A) / Length(B)); }

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

 Vector VRotate(Vector A, double rad)
 {
     return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
 }

 Vector Normal(Vector A)
 {
     double l = Length(A);
     return Vector(-A.y/l, A.x/l);
 }

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

 double Cross(Vector A, Vector B)
 { return A.x*B.y - A.y*B.x; }

 struct Circle
 {
     double x, y, r;
     Circle(double x=, double y=, double r=):x(x), y(y), r(r) {}
     Point point(double a)
     {
         return Point(x+r*cos(a), y+r*sin(a));
     }
 };

 const int maxn = ;
 char s[maxn];

 int ID(char* s)
 {
     int l = strlen(s);
     switch(l)
     {
         case : return ;
         case : return ;
         case : return ;
         case : return ;
         case : return ;
         case : return ;
         default: return -;
     }
 }

 void Solve(double A1, double B1, double C1, double A2, double B2, double C2, double& ansx, double& ansy)
 {
     ansx = (B1*C2 - B2*C1) / (A1*B2 - A2*B1);
     ansy = (C2*A1 - C1*A2) / (B1*A2 - B2*A1);
 }

 void problem0()
 {
     Point A, B, C;
     scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y);
     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)/;
     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)/;
     Point ans;
     Solve(A1, B1, C1, A2, B2, C2, ans.x, ans.y);
     double r = Length(ans - A);
     printf("(%.6lf,%.6lf,%.6lf)\n", ans.x, ans.y, r);
 }

 void problem1()
 {
     Point A, B, C;
     scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y);
     double a = Length(B-C), b = Length(A-C), c = Length(A-B);
     double l = a+b+c;
     Point ans = (A*a+B*b+C*c)/l;
     double r = fabs(Cross(A-B, C-B)) / l;
     printf("(%.6lf,%.6lf,%.6lf)\n", ans.x, ans.y, r);
 }

 void problem2()
 {
     Circle C;
     Point P, O;
     scanf("%lf%lf%lf%lf%lf", &C.x, &C.y, &C.r, &P.x, &P.y);
     double ans[];
     O.x = C.x, O.y = C.y;
     double d = Length(P-O);
     int k = dcmp(d-C.r);
     if(k < )
     {
         printf("[]\n");
         return;
     }
     else if(k == )
     {
         ans[] = Change(Angle2(P-O)) + 90.0;
         while(ans[] >= 180.0)    ans[] -= 180.0;
         while(ans[] < )        ans[] += 180.0;
         printf("[%.6lf]\n", ans[]);
         return;
     }
     else
     {
         double ag = asin(C.r/d);
         double base = Angle2(P-O);
         ans[] = base + ag, ans[] = base - ag;
         ans[] = Change(ans[]), ans[] = Change(ans[]);
         while(ans[] >= 180.0)    ans[] -= 180.0;
         while(ans[] < )        ans[] += 180.0;
         while(ans[] >= 180.0)    ans[] -= 180.0;
         while(ans[] < )        ans[] += 180.0;
         if(ans[] >= ans[])    swap(ans[], ans[]);
         printf("[%.6lf,%.6lf]\n", ans[], ans[]);
     }
 }

 vector<Point> sol;
 struct Line
 {
     Point p;
     Vector v;
     Line()    { }
     Line(Point p, Vector v): p(p), v(v)    {}
     Point point(double t)
     {
         return p + v*t;
     }
     Line move(double d)
     {
         return Line(p + Normal(v)*d, v);
     }
 };
 Point GetIntersection(Line a, Line b)
 {
     Vector u = a.p - b.p;
     double t = Cross(b.v, u) / Cross(a.v, b.v);
     return a.p + a.v*t;
 }
 struct Circle2
 {
     Point c;    //圆心
     double r;    //半径
     Point point(double a)
     {
         return Point(c.x+r*cos(a), c.y+r*sin(a));
     }
 };
 //两圆相交并返回交点个数
 int getLineCircleIntersection(Line L, Circle2 C, vector<Point>& sol)
 {
     double t1, t2;
     double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
     double e = a*a + c*c, f = *(a*b + c*d), g = b*b + d*d - C.r*C.r;
     double delta = f*f - *e*g;        //判别式
     if(dcmp(delta) < )    return ;    //相离
     if(dcmp(delta) == )            //相切
     {
         t1 = t2 = -f / ( * e);
         sol.push_back(L.point(t1));
         return ;
     }
     //相交
     t1 = (-f - sqrt(delta)) / ( * e);    sol.push_back(L.point(t1));
     t2 = (-f + sqrt(delta)) / ( * e);    sol.push_back(L.point(t2));
     return ;
 }
 void problem3()
 {
     Circle2 C;
     Point A, B;
     scanf("%lf%lf%lf%lf%lf%lf%lf", &C.c.x, &C.c.y, &A.x, &A.y, &B.x, &B.y, &C.r);
     Vector v = (A-B)/Length(A-B)*C.r;
     //printf("%lf\n", Length(v));
     Point p1 = A + Point(-v.y, v.x);
     Point p2 = A + Point(v.y, -v.x);
     //printf("%lf\n%lf", Length(p1-C.c), Length(p2-C.c));
     Line L1(p1, v), L2(p2, v);

     sol.clear();
     int cnt =  getLineCircleIntersection(L1, C, sol);
     cnt += getLineCircleIntersection(L2, C, sol);
     sort(sol.begin(), sol.end());
     if(cnt == )    { printf("[]\n"); return; }
     printf("[");
     for(int i = ; i < cnt-; ++i)    printf("(%.6lf,%.6lf),", sol[i].x, sol[i].y);
     printf("(%.6lf,%.6lf)]\n", sol[cnt-].x, sol[cnt-].y);
 }

 void problem4()
 {
     double r;
     Point A, B, C, D, E, ans[];
     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);
     Line a(A, B-A), b(C, D-C);
     Line L1 = a.move(r), L2 = a.move(-r);
     Line L3 = b.move(r), L4 = b.move(-r);
     ans[] = GetIntersection(L1, L3);
     ans[] = GetIntersection(L1, L4);
     ans[] = GetIntersection(L2, L3);
     ans[] = GetIntersection(L2, L4);
     sort(ans, ans+);
     printf("[");
     for(int i = ; i < ; ++i)    printf("(%.6lf,%.6lf),", ans[i].x, ans[i].y);
     printf("(%.6lf,%.6lf)]\n", ans[].x, ans[].y);
 }

 int getCircleCircleIntersection(Circle2 C1, Circle2 C2, vector<Point>& sol)
 {
     double d = Length(C1.c - C2.c);
     if(dcmp(d) == )
     {
         if(dcmp(C1.r - C2.r) == )    return -;
         return ;
     }
     if(dcmp(C1.r + C2.r - d) < )    return ;
     if(dcmp(fabs(C1.r - C2.r) - d) > )    return ;

     double a = Angle2(C2.c - C1.c);
     double da = acos((C1.r*C1.r + d*d - C2.r*C2.r) / (*C1.r*d));
     Point p1 = C1.point(a+da), p2 = C1.point(a-da);
     sol.push_back(p1);
     if(p1 == p2)    return ;
     sol.push_back(p2);
     return ;
 }

 void problem5()
 {
     Circle2 C1, C2;
     double r;
     vector<Point> sol;
     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);
     double d = Length(C1.c - C2.c);
     C1.r += r, C2.r += r;
     if(dcmp(C1.r+C2.r-d) < )    { printf("[]\n"); return; }
     int n = getCircleCircleIntersection(C1, C2, sol);
     sort(sol.begin(), sol.end());
     printf("[");
     for(int i = ; i < n-; ++i)    printf("(%.6lf,%.6lf),", sol[i].x, sol[i].y);
     printf("(%.6lf,%.6lf)]\n", sol[n-].x, sol[n-].y);
 }

 int main()
 {
     #ifdef    LOCAL
         freopen("12304in.txt", "r", stdin);
     #endif

     while(scanf("%s", s) == )
     {
         int proID = ID(s);
         switch(proID)
         {
             case :    problem0();    break;
             case :    problem1();    break;
             case :    problem2();    break;
             case :    problem3();    break;
             case :    problem4();    break;
             case :    problem5();    break;
             default: break;
         }
     }

     return ;
 }

代码君