LightOJ 1331（数学计算几何）

最新推荐文章于 2017-07-20 18:32:00 发布

原创最新推荐文章于 2017-07-20 18:32:00 发布 · 524 阅读

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本文介绍了一个几何问题的解决方法，即如何计算三个相切圆形覆盖区域内的特定蓝色阴影区域的面积。通过构建圆心三角形并利用三角形的边长信息，文章详细解释了如何计算该三角形的面积及阴影部分的面积。

Description
Agent J is preparing to steal an antique diamond piece from a museum. As it is fully guarded and they are guarding it using high technologies, it’s not easy to steal the piece. There are three circular laser scanners in the museum which are the main headache for Agent J. The scanners are centered in a certain position, and they keep rotating maintaining a certain radius. And they are placed such that their coverage areas touch each other as shown in the picture below:

这里写图片描述

Here R1, R2 and R3 are the radii of the coverage areas of the three laser scanners. The diamond is placed in the place blue shaded region as in the picture. Now your task is to find the area of this region for Agent J, as he needs to know where he should land to steal the diamond.

Input
Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing three real numbers denoting R1, R2 and R3 (0 < R1, R2, R3 ≤ 100). And no number contains more than two digits after the decimal point.

Output
For each case, print the case number and the area of the place where the diamond piece is located. Error less than 10-6 will be ignored.

Sample Input
3
1.0 1.0 1.0
2 2 2
3 3 3
Sample Output
Case 1: 0.16125448
Case 2: 0.645017923
Case 3: 1.4512903270

题意：给出3个圆的半径，让你求阴影部分的面积。

可以连接三个圆的圆心，可以构成一个三角形，并且三角形的三条边都知道了，这是三角形的面积也就有了，利用p=（a+b+c）/2,可求。

知道三边也就知道

#include<cstdio>
#include<cmath>
#define PI acos(-1.0)
int main()
{
	int t,flag=1;
	scanf("%d",&t);
	while(t--)
	{
		double r1,r2,r3;
		scanf("%lf%lf%lf",&r1,&r2,&r3);
		double a,b,c,cosa,cosb,cosc;
		a=r1+r2;
		b=r2+r3;
		c=r3+r1;
		double p=(a+b+c)/2.0;
		double s=sqrt(p*(p-a)*(p-b)*(p-c)); 
       	cosa=(b*b+c*c-a*a)/(2.0*b*c);//求余弦 
		cosb=(a*a+c*c-b*b)/(2.0*a*c);
		cosc=(b*b+a*a-c*c)/(2.0*b*a);
		double s1=r3*r3*acos(cosa)/2.0+r2*r2*acos(cosc)/2.0+r1*r1*acos(cosb)/2.0;
		printf("Case %d: %lf\n",flag++,s-s1);	
	} 
	return 0;
}

定理，就知道角度了，acos（余弦）就ok了。