hdu6354 Everything Has Changed 几何题

最新推荐文章于 2020-03-08 19:50:52 发布

Hugo5332

最新推荐文章于 2020-03-08 19:50:52 发布

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分类专栏： ACM ACM算法

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本文介绍了一种计算在二维平面上，经过多个圆形切割区域后，初始圆形模型剩余部分周长的方法。通过判断小圆与大圆的位置关系，采用不同的计算策略，特别是对于相交情况利用余弦定理求解角度，进而计算弧长。

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Everything Has Changed

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Special Judge

Problem Description

Edward is a worker for Aluminum Cyclic Machinery. His work is operating mechanical arms to cut out designed models. Here is a brief introduction of his work.
Assume the operating plane as a two-dimensional coordinate system. At first, there is a disc with center coordinates (0,0) and radius R. Then, m mechanical arms will cut and erase everything within its area of influence simultaneously, the i-th area of which is a circle with center coordinates (xi,yi) and radius ri (i=1,2,⋯,m). In order to obtain considerable models, it is guaranteed that every two cutting areas have no intersection and no cutting area contains the whole disc.
Your task is to determine the perimeter of the remaining area of the disc excluding internal perimeter.
Here is an illustration of the sample, in which the red curve is counted but the green curve is not.

Input

The first line contains one integer T, indicating the number of test cases.
The following lines describe all the test cases. For each test case:
The first line contains two integers m and R.
The i-th line of the following m lines contains three integers xi,yi and ri, indicating a cutting area.
1≤T≤1000, 1≤m≤100, −1000≤xi,yi≤1000, 1≤R,ri≤1000 (i=1,2,⋯,m).

Output

For each test case, print the perimeter of the remaining area in one line. Your answer is considered correct if its absolute or relative error does not exceed 10−6.
Formally, let your answer be a and the jury's answer be b. Your answer is considered correct if |a−b|max(1,|b|)≤10−6.

Sample Input

1 4 10 6 3 5 10 -4 3 -2 -4 4 0 9 1

Sample Output

81.62198908430238475376

解题思路 : 有三种情况小圆在大圆外和大圆内不用考虑内切的话直接加上小圆周长相交的情况先画个图发现可以通过两次余弦定理求出角度来求出弧长，相加即可。

#include<bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
struct cir {
	int x;
	int r;
	int y;
} a[50000];
double _distance(cir a,cir b) {
	return 1.0*sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double _cos(double a,double b,double c) {
	return 1.0*(a*a+b*b-c*c)/(2*a*b);
}
int N;
double R;
int main() {
	int t;
	cin>>t;
	while(t--) {
		cin>>N>>R;
		cir m;
		m.x=m.y=0;
		m.r=R;
		//cout<<PI<<endl;
		double ans=2*PI*R;
		for(int i=0; i<N; i++) scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].r);
		for(int i=0; i<N; i++) {
			if(_distance(a[i],m)+a[i].r<R||_distance(a[i],m)>=R+a[i].r) {
				// cout<<_distance(a[i],m)<<" "<<a[i].r<<endl;
				continue;
			}
			if(fabs(_distance(a[i],m)+a[i].r-R)<1e-6) {
				ans+=(2*PI*a[i].r);
			} else {
				// cout<<"22"<<endl;
				double r1=_distance(a[i],m);
				double cos_x = _cos(R,r1,a[i].r);
				double x_h = acos(cos_x);
				x_h*=2;
				x_h*=180;
				x_h/=PI;
				double h_R = (1.0*x_h/360)*(2*PI*R);

				double cos_y =_cos(r1,a[i].r,R);
				double y_h = acos(cos_y);
				y_h*=2;
				y_h*=180;
				y_h/=PI;
				double h_r = (1.0*y_h/360)*(2*PI*a[i].r);
				ans+=(h_r-h_R);

			}
		}

		printf("%.8lf\n",ans);


	}
}