Codeforces Beta Round #2 C. Commentator problem 模拟退火

C. Commentator problem

题目连接：

http://www.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

Hint

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ang = asin(D/r)这个很显然。

然后我们确保ang都相同就好了。

然后角度越大，其实就是距离越近。

然后这个东西直接退火一下就好了。

嘟嘟嘟，随便退一退。

代码

#include<bits/stdc++.h>
using namespace std;
struct node
{
    double x,y;
    double r;
}p[3];
double dis(double x,double y,node t)
{
    return sqrt((x-t.x)*(x-t.x)+(y-t.y)*(y-t.y));
}
double cost(double x,double y)
{
    double ang[3];
    for(int i=0;i<3;i++)
        ang[i]=dis(x,y,p[i])/p[i].r;
    double d[3];
    for(int i=0;i<3;i++)
        d[i]=ang[i]-ang[(i+1)%3];
    return d[0]*d[0]+d[1]*d[1]+d[2]*d[2];
}
int main()
{
    for(int i=0;i<3;i++)
        cin>>p[i].x>>p[i].y>>p[i].r;
    double x=0,y=0;
    for(int i=0;i<3;i++)
        x+=p[i].x/3,y+=p[i].y/3;
    double t=1.0;
    while(t>1e-5)
    {
        int flag = 0;
        if(cost(x+t,y)<cost(x,y))x+=t,flag=1;
        else if(cost(x,y+t)<cost(x,y))y+=t,flag=1;
        else if(cost(x-t,y)<cost(x,y))x-=t,flag=1;
        else if(cost(x,y-t)<cost(x,y))y-=t,flag=1;
        if(!flag)t*=0.5;
    }
    if(fabs(cost(x,y))<1e-5)printf("%.12f %.12f
",x,y);
}