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UVa 10674 (求两圆公切线) Tangents

题意:

给出两个圆的圆心坐标和半径,求这两个圆的公切线切点的坐标及对应线段长度。若两圆重合,有无数条公切线则输出-1.

输出是按照一定顺序输出的。

分析:

首先情况比较多,要一一判断,不要漏掉。

如果高中的那点老底还在的话,代码还是很好理解的。

 

  1 //#define LOCAL  2 #include <cstdio>  3 #include <cstring>  4 #include <algorithm>  5 #include <cmath>  6 #include <vector>  7 using namespace std;  8   9 const double PI = acos(-1.0); 10 const double EPS = 1e-8; 11 struct Point 12 { 13     double x, y; 14     Point(double x=0, double y=0):x(x), y(y) {} 15  16 }; 17 typedef Point Vector; 18 Vector operator + (Vector A, Vector B) 19 { 20     return Vector(A.x+B.x, A.y+B.y); 21 } 22 Vector operator - (Vector A, Vector B) 23 { 24     return Vector(A.x-B.x, A.y-B.y); 25 } 26 Vector operator * (Vector A, double p) 27 { 28     return Vector(A.x*p, A.y*p); 29 } 30 Vector operator / (Vector A, double p) 31 { 32     return Vector(A.x/p, A.y/p); 33 } 34 double dcmp(double x) 35 { 36     if(fabs(x) < EPS)    return 0; 37     else return x < 0 ? -1 : 1; 38 } 39 bool operator < (const Vector& a, const Vector& b) 40 { 41     return dcmp(a.x-b.x) < 0 || dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) < 0; 42 } 43 bool operator == (const Vector& a, const Vector& b) 44 { 45     return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; 46 } 47 double Dot(Vector a, Vector b) 48 { 49     return a.x*b.x + a.y*b.y; 50 } 51 double Cross(Vector a, Vector b) 52 { 53     return a.x*b.y - a.y*b.x; 54 } 55 double Length(Vector a) 56 { 57     return sqrt(Dot(a, a)); 58 } 59 struct Circle 60 { 61     double x, y, r; 62     Circle(double x, double y, double r):x(x), y(y), r(r) {} 63     Point point(double a) 64     { 65         return Point(x + r*cos(a), y + r*sin(a)); 66     } 67 }; 68 int getTangents(Circle A, Circle B, Point* a, Point* b) 69 { 70     int cnt = 0; 71     if(A.r < B.r)    { swap(A, B); swap(a, b); } 72     double d2 = (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y); 73     double rdiff = A.r - B.r; 74     double rsum = A.r + B.r; 75     if(d2 < rdiff*rdiff)    return 0;    //内含 76  77     double base = atan2(B.y-A.y, B.x-A.x); 78     if(dcmp(d2) == 0 && dcmp(A.r - B.r) == 0)    return -1; //重合 79     if(dcmp(d2 - rdiff*rdiff) == 0)    //内切 80     { 81         a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++; 82         return 1; 83     } 84  85     //有外公切线 86     double ang = acos((A.r - B.r) / sqrt(d2)); 87     a[cnt] = A.point(base + ang); b[cnt] = B.point(base + ang); cnt++; 88     a[cnt] = A.point(base - ang); b[cnt] = B.point(base - ang); cnt++; 89     if(dcmp(rsum*rsum - d2) == 0) 90     {//外切 91         a[cnt] = b[cnt] = A.point(base); cnt++; 92     } 93     else if(dcmp(d2 - rsum*rsum) > 0) 94     { 95         ang = acos((A.r + B.r) / sqrt(d2)); 96         a[cnt] = A.point(base + ang); b[cnt] = B.point(PI + base + ang); cnt++; 97         a[cnt] = A.point(base - ang); b[cnt] = B.point(PI + base - ang); cnt++; 98     } 99     return cnt;100 }101 102 int main(void)103 {104     #ifdef    LOCAL105         freopen("10674in.txt", "r", stdin);106     #endif107     108     int x1, y1, r1, x2, y2, r2;109     while(scanf("%d%d%d%d%d%d", &x1, &y1, &r1, &x2, &y2, &r2) == 6 && r1 && r2)110     {111         Point a[4], b[4];112         Circle C1(x1, y1, r1), C2(x2, y2, r2);113         int n = getTangents(C1, C2, a, b);114         printf("%d\n", n);115         int p[4] = {0, 1, 2, 3};116         for(int i = 0; i < n; ++i)117             for(int j = i+1; j < n; ++j)118                 if(a[p[j]] < a[p[i]] || (a[p[j]] == a[p[i]] && b[p[j]] < b[p[i]]))    swap(p[i], p[j]);119         for(int i = 0; i < n; ++i)120             printf("%.5lf %.5lf %.5lf %.5lf %.5lf\n", a[p[i]].x, a[p[i]].y, b[p[i]].x, b[p[i]].y, Length(a[p[i]] - b[p[i]]));121     }122 123     return 0;124 }
代码君

 

写的第一份,样例过,自己随机生成的1000组数据和lrj的代码对比,输出也是一模一样的,可就是无线WA。

这里还是贴一下这份“神奇”的代码吧,也许日后能发现这个隐藏的错误。

  1 //#define LOCAL  2 #include <cstdio>  3 #include <cstring>  4 #include <algorithm>  5 #include <vector>  6 #include <cmath>  7 using namespace std;  8   9 struct Point 10 { 11     double x, y; 12     Point(double x=0, double y=0) :x(x),y(y) {} 13 }; 14 typedef Point Vector; 15 const double EPS = 1e-8; 16 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); } 17 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); } 18 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); } 19 Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); } 20 bool operator < (const Point& a, const Point& b) 21 { return a.x < b.x || (a.x == b.x && a.y < b.y); } 22 int dcmp(double x) 23 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; } 24 bool operator == (const Point& a, const Point& b) 25 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; } 26 double Dot(Vector A, Vector B) 27 { return A.x*B.x + A.y*B.y; } 28 double Length(Vector A)    { return sqrt(Dot(A, A)); } 29  30 struct Circle 31 { 32     Point c;    //Ô²ÐÄ 33     double r;    //°ë¾¶ 34     //Circle(Point c, double r):c(c), r(r)    {} 35     Point point(double a) 36     {//Çó¶ÔÓ¦Ô²ÐĽǵĵã 37         return Point(c.x + r*cos(a), c.y + r*sin(a)); 38     } 39 }; 40 double angle(Vector v)    { return atan2(v.y, v.x); } 41 const double PI = acos(-1.0);  42  43 int getTangents(Circle A, Circle B, Point* a, Point* b) 44 { 45     int cnt = 0; 46     if(A.r < B.r)    { swap(A, B); swap(a, b); } 47     double d2 = (A.c.x-B.c.x)*(A.c.x-B.c.x) + (A.c.y-B.c.y)*(A.c.y-B.c.y); 48     double rdiff = A.r - B.r; 49     double rsum  = A.r + B.r; 50     if(d2 < rdiff*rdiff)    return 0;    //ÄÚº¬ 51  52     double base = atan2(B.c.y-A.c.y, B.c.x-A.c.x); 53     if(dcmp(d2) == 0 && dcmp(A.r-B.r) == 0)    return -1;    //Á½Ô²Öغϣ¬ÎÞÇî¶àÌõÇÐÏß 54     if(dcmp(d2 - rdiff*rdiff) == 0)                //ÄÚÇÐ 55     { 56         a[cnt] = A.point(base); 57         b[cnt] = B.point(base); 58         cnt++; 59         return 1; 60     } 61     //ÓÐÍ⹫ÇÐÏß 62     double ang = acos((A.r-B.r) / sqrt(d2)); 63     a[cnt] = A.point(base+ang); b[cnt] = B.point(base+ang); cnt++; 64     a[cnt] = A.point(base-ang); b[cnt] = B.point(base-ang); cnt++; 65      66     if(dcmp(d2 - rsum*rsum) == 0)    //Ò»ÌõÄÚ¹«ÇÐÏß 67     { 68         a[cnt] = b[cnt] = A.point(base); cnt++; 69     } 70     else if(dcmp(d2 - rsum*rsum) > 0) 71     { 72         double ang = acos((A.r+B.r) / sqrt(d2)); 73         a[cnt] = A.point(base+ang); b[cnt] = B.point(base+ang+PI); cnt++; 74         a[cnt] = A.point(base-ang); b[cnt] = B.point(base-ang+PI); cnt++; 75     } 76     //for(int i = 0; i < cnt; ++i) 77       //    printf("%.5lf %.5lf %.5lf %.5lf\n", a[i].x, a[i].y, b[i].x, b[i].y); 78     return cnt; 79 } 80  81 int main(void) 82 { 83     #ifdef    LOCAL 84         freopen("10674in.txt", "r", stdin); 85         //freopen("10674out.txt", "w", stdout); 86     #endif 87  88     int x1, y1, r1, x2, y2, r2; 89     while(scanf("%d%d%d%d%d%d", &x1, &y1, &r1, &x2, &y2, &r2) == 6 && dcmp(r1) > 0) 90     { 91         Point a[10], b[10]; 92         /*Point p1(x1, y1), p2(x2, y2); 93         Circle C1(p1, r1), C2(p2, r2);*/ 94         Circle C1, C2; 95         C1.c.x=x1, C1.c.y=y1, C1.r=r1; 96         C2.c.x=x2, C2.c.y=y2, C2.r=r2; 97         int n = getTangents(C1, C2, a, b); 98         printf("%d\n", n); 99         if(n > 0)100         {101             int p[4] = {0, 1, 2, 3};102             for(int i = 0; i < n; ++i)103                 for(int j = i+1; j < n; ++j)104                     if(a[p[j]] < a[p[i]] || (a[p[j]] == a[p[i]] && b[p[j]] < b[p[i]]))    swap(p[i], p[j]);105             for(int i = 0; i < n; ++i)106                 printf("%.5lf %.5lf %.5lf %.5lf %.5lf\n", a[p[i]].x, a[p[i]].y, b[p[i]].x, b[p[i]].y, Length(a[p[i]] - b[p[i]]));107         }108     }109 110     return 0;111 }
WA掉的代码君

 

UVa 10674 (求两圆公切线) Tangents