After so many years of competition your friend Marcos has decided to quit competitive programming, he is tired of solving problems. In this radical change of life he has decided to open a little newsstand in the park where he can enjoy a quieter life.
The problem is that the park where he wants to work has 3 fountains that are active all afternoons. Marcos doesn't want his customers to become wet while chatting with them as they choose their newspaper, so he's looking for the perfect place to set up his store.
The amount of water that reaches each point of the park is defined by the fountain closest to it. It is true that the rest of the fountains also wet, but it is so little that it can be ignored. Therefore, Mark's objective is to place his tent in such a way as to maximize the distance to the nearest fountain, i.e., to maximize the value of $$$\texttt{DRYNESS}$$$ described by: $$$$$$\texttt{DRYNESS}(p) =\min \{ d(p,F_1), d(p,F_2), d(p,F_3) \}$$$$$$ where $$$F_1$$$,$$$F_2$$$ y $$$F_3$$$ represent the positions of the fountains and the function $$$d(p, q)$$$ represents the Euclidean distance in the plane between two points $$$p$$$ and $$$q$$$, that is, the usual distance in a straight line.
Knowing that the park enclosure is a perfect square of side $$$100$$$ and its lower left corner is at the point $$$(0,0)$$$, can you indicate what is the largest value of $$$\texttt{DRYNESS}$$$ that Marcos will be able to achieve by placing his new store?
The input starts with a number $$$N$$$, the number of test cases to be processed. Each test case is composed of 3 lines, each indicating a pair of numbers $$$(F_x,F_y)$$$, the position of one of the fountains. All these values will be indicated with 3 decimal places, using the point as decimal separator.
The output for each case will indicate a three digits rounded number (using the point as decimal separator), the maximum value of DRYNESS that Marcos can achieve by placing his tent in the park.
1 19.000 13.000 10.000 81.000 73.000 44.000
62.169
To print a number $$$x$$$ with 3 decimal digits, you can use:
std::cout « std::fixed « std::setprecision(3) « x « std::endl;
double round = Math.round(x * 1000.0) / 1000.0;
String numberRounded = String.format(" System.out.println(numberRounded);
print(f'{x:.3f}')