poj1329Circle Through Three Points(三角形外接圆）

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<stdlib.h>
 6 #include<vector>
 7 #include<cmath>
 8 #include<queue>
 9 #include<set>
10 using namespace std;
11 #define N 100000
12 #define LL long long
13 #define INF 0xfffffff
14 const double eps = 1e-8;
15 const double pi = acos(-1.0);
16 const double inf = ~0u>>2;
17 struct Point
18 {
19     double x,y;
20     Point (double x=0,double y =0):x(x),y(y){}
21 }p[4];
22 struct Circle
23 {
24     Point center;
25     double r;
26 };
27 typedef Point pointt;
28 pointt operator - (Point a,Point b)
29 {
30     return Point(a.x-b.x,a.y-b.y);
31 }
32 int dcmp(double x)
33 {
34     if(fabs(x)<eps) return 0;
35     return x<0?-1:1;
36 }
37 double dis(Point a)
38 {
39     return a.x*a.x+a.y*a.y;
40 }
41 double cross(Point a,Point b)
42 {
43     return a.x*b.y-a.y*b.x;
44 }
45 double area()
46 {
47     return fabs(cross(p[1]-p[2],p[2]-p[3]))/2;
48 }
49 struct Circle Circumcircle()
50 {
51     Circle tmp;
52     double a,b,c,c1,c2;
53     double xa,ya,xb,yb,xc,yc;
54     a = sqrt(dis(p[3]-p[1]));
55     b = sqrt(dis(p[1]-p[2]));
56     c = sqrt(dis(p[2]-p[3]));
57     //根据s = a*b*c/R/4，求半径
58     tmp.r = (a*b*c)/(area()*4.0);
59     xa = p[3].x;
60     ya = p[3].y;
61     xb = p[1].x;
62     yb = p[1].y;
63     xc = p[2].x;
64     yc = p[2].y;
65     c1 = (dis(p[3])-dis(p[1]))/2;
66     c2 = (dis(p[3])-dis(p[2]))/2;
67     tmp.center.x = (c1*(ya-yc)-c2*(ya-yb))/((xa-xb)*(ya-yc)-(xa-xc)*(ya-yb));
68     tmp.center.y = (c1*(xa-xc)-c2*(xa-xb))/((ya-yb)*(xa-xc)-(ya-yc)*(xa-xb));
69     return tmp;
70 }
71 int main()
72 {
73     while(scanf("%lf%lf%lf%lf%lf%lf",&p[1].x,&p[1].y,&p[2].x,&p[2].y,&p[3].x,&p[3].y)!=EOF)
74     {
75         Circle cc = Circumcircle();
76         double r = cc.r;
77         char c1,c2;
78         c1 = dcmp(cc.center.x)>0?‘-‘:‘+‘;
79         c2 = dcmp(cc.center.y)>0?‘-‘:‘+‘;
80         printf("(x %c %.3f)^2 + (y %c %.3f)^2 = %.3f^2\n",c1,fabs(cc.center.x),c2,fabs(cc.center.y),r);
81         double s = r*r-cc.center.x*cc.center.x-cc.center.y*cc.center.y;
82         char c3 = dcmp(s)>0?‘-‘:‘+‘;
83         printf("x^2 + y^2 %c %.3fx %c %.3fy %c %.3f = 0\n",c1,fabs(cc.center.x*2),c2,fabs(cc.center.y*2),c3,fabs(s));
84         puts("");
85     }
86     return 0;
87 }