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UVa 10674 (求两圆公切线) Tangents

时间:2014-10-16 02:21:51      阅读:133      评论:0      收藏:0      [点我收藏+]

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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。

 

bubuko.com,布布扣
  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    LOCAL
105         freopen("10674in.txt", "r", stdin);
106     #endif
107     
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。

bubuko.com,布布扣
  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

标签:style   blog   http   color   io   os   ar   for   sp   

原文地址:http://www.cnblogs.com/AOQNRMGYXLMV/p/4027743.html

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