C. Commentator problem （模拟退火）

左手的ㄟ右手 2022-06-18 08:40 131阅读 0赞

[点击打开链接][Link 1]

http://codeforces.com/contest/2/problem/C

## Description ##

The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered.

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

## Input ##

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x,  y,  r, where (x, y) are the coordinates of the stadium's center ( -  103 ≤ x,  y ≤ 103), and r (1 ≤ r  ≤ 103) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line.

## Output ##

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

## Sample Input ##

0 0 10  
60 0 10  
30 30 10

## Sample Output ##

30.00000 0.00000

题意：

给你三个圆，现在你要找一个点，使得这个点和每个圆形成的切线的角度都是一样的。  
如果有多个点，那么选择角度最大的那个点。

题解：

假设离那个圆的圆心的距离为d，那么angle = asin(d/r)这个很显然。  
那么我们确保angle都相同就好了。  
然后角度越大，其实就是距离越近。  
最后把这个东西直接模拟退火就好了。

AC代码：

#include<bits/stdc++.h>
    using namespace std;
    double x[3],y[3],z[3];
    double t[3];
    double cal(double x)
    {
    	return x*x;
    }
    double check(double tx,double ty)
    {
    	double ans=0;
    	for(int i=0;i<3;i++)
    	{
    		t[i] = (double)sqrt(cal(tx-x[i]) + cal(ty-y[i])) / z[i]; 
    	}
    	for(int i=0;i<3;i++) 
    	{
    		ans += cal(t[i]-t[(i+1)%3]);
    	}
    	return ans;
    }
    int main()
    {
    	//freopen("in.txt","r",stdin);
    	for(int i=0;i<3;i++){
    		scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
    	}
    	double mx=(x[0] + x[1] + x[2])/3.0;
    	double my=(y[0] + y[1] + y[2])/3.0;
    	int flag=0;
    	double d=1.0;
    	while(d>1e-5) 
    	{
        	flag=0;
    		double tmp = check(mx,my);
    		if(tmp > check(mx+d,my)) mx += d, flag=1;
    		else if(tmp > check(mx-d,my)) mx -= d,flag=1;
    		else if(tmp > check(mx,my+d)) my += d,flag=1;
    		else if(tmp > check(mx,my-d)) my -= d ,flag=1;
    		if(flag==0) d *= 0.7;
    	}
    	if(check(mx,my)< 1e-5) printf("%.5f %.5f\n",mx,my);
    	return 0;
    }

[Link 1]: http://codeforces.com/contest/2/problem/C