首页 > 代码库 > hdu 5120 Intersection
hdu 5120 Intersection
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5120
A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below.
Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.
题意：求两个相同大小的圆环的面积交。
解法：几何，圆的面积交模板。画一个图应该就能发现公式：圆环面积交=大圆大圆面积交-2×大圆小圆面积交+小圆小圆面积交。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #include<algorithm> 7 #define inf 0x7fffffff 8 #define exp 1e-10 9 #define PI 3.14159265410 using namespace std;11 typedef long long ll;12 struct Point13 {14 double x,y;15 Point (double x=0,double y=0):x(x),y(y){}16 };17 double dist(Point a,Point b)18 {19 double x=(a.x-b.x)*(a.x-b.x);20 double y=(a.y-b.y)*(a.y-b.y);21 return sqrt(x+y);22 }23 double Area_of_overlap(Point c1,double r1,Point c2,double r2)24 {25 double d=dist(c1,c2);26 if (r1+r2<d+exp) return 0;27 if (d<fabs(r1-r2)+exp)28 {29 double r=min(r1,r2);30 return PI*r*r;31 }32 double x=(d*d+r1*r1-r2*r2)/(2*d);33 double t1=acos(x/r1);34 double t2=acos((d-x)/r2);35 return r1*r1*t1+r2*r2*t2-d*r1*sin(t1);36 }37 int main()38 {39 int t,ncase=1;40 double r,R;41 Point a,b;42 scanf("%d",&t);43 while (t--)44 {45 scanf("%lf%lf",&r,&R);46 scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);47 double bb_area=Area_of_overlap(a,R,b,R);48 double bs_area=Area_of_overlap(a,R,b,r);49 double ss_area=Area_of_overlap(a,r,b,r);50 printf("Case #%d: %.6lf\n",ncase++,bb_area-2.0*bs_area+ss_area);51 }52 return 0;53 }
hdu 5120 Intersection
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。