Balanced Removals (Easier)
Contest: Codeforces Global Round 5
Difficulty rating: 1700
Tags: constructive algorithms, geometry, greedy

This is an easier version of the problem. In this version, $$$n \le 2000$$$.

There are $$$n$$$ distinct points in three-dimensional space numbered from $$$1$$$ to $$$n$$$. The $$$i$$$-th point has coordinates $$$(x_i, y_i, z_i)$$$. The number of points $$$n$$$ is even.

You'd like to remove all $$$n$$$ points using a sequence of $$$\frac{n}{2}$$$ snaps. In one snap, you can remove any two points $$$a$$$ and $$$b$$$ that have not been removed yet and form a perfectly balanced pair. A pair of points $$$a$$$ and $$$b$$$ is perfectly balanced if no other point $$$c$$$ (that has not been removed yet) lies within the axis-aligned minimum bounding box of points $$$a$$$ and $$$b$$$.

Formally, point $$$c$$$ lies within the axis-aligned minimum bounding box of points $$$a$$$ and $$$b$$$ if and only if $$$\min(x_a, x_b) \le x_c \le \max(x_a, x_b)$$$, $$$\min(y_a, y_b) \le y_c \le \max(y_a, y_b)$$$, and $$$\min(z_a, z_b) \le z_c \le \max(z_a, z_b)$$$. Note that the bounding box might be degenerate.

Find a way to remove all points in $$$\frac{n}{2}$$$ snaps.

Input format:
The first line contains a single integer $$$n$$$ ($$$2 \le n \le 2000$$$; $$$n$$$ is even), denoting the number of points.

Each of the next $$$n$$$ lines contains three integers $$$x_i$$$, $$$y_i$$$, $$$z_i$$$ ($$$-10^8 \le x_i, y_i, z_i \le 10^8$$$), denoting the coordinates of the $$$i$$$-th point.

No two points coincide.

Output format:
Output $$$\frac{n}{2}$$$ pairs of integers $$$a_i, b_i$$$ ($$$1 \le a_i, b_i \le n$$$), denoting the indices of points removed on snap $$$i$$$. Every integer between $$$1$$$ and $$$n$$$, inclusive, must appear in your output exactly once.

We can show that it is always possible to remove all points. If there are many solutions, output any of them.

Constraints:
Time limit: 1.0 s
Memory limit: 512.0 MB

Note:
In the first example, here is what points and their corresponding bounding boxes look like (drawn in two dimensions for simplicity, as all points lie on $$$z = 0$$$ plane). Note that order of removing matters: for example, points $$$5$$$ and $$$1$$$ don't form a perfectly balanced pair initially, but they do after point $$$3$$$ is removed.

Example input:
6
3 1 0
0 3 0
2 2 0
1 0 0
1 3 0
0 1 0

Example output:
3 6
5 1
2 4

Example input:
8
0 1 1
1 0 1
1 1 0
1 1 1
2 2 2
3 2 2
2 3 2
2 2 3

Example output:
4 5
1 6
2 7
3 8

Editorial:
Pick any two points $$$i$$$ and $$$j$$$, let's say that this is our candidate pair $$$(i, j)$$$ for removal. Loop over all other points. If some point $$$k$$$ lies inside the bounding box of $$$i$$$ and $$$j$$$, change our candidate pair to $$$(i, k)$$$. Note that the bounding box of $$$i$$$ and $$$k$$$ lies inside the bounding box of $$$i$$$ and $$$j$$$, so we don't need to recheck points that we have already checked. The candidate pair we get at the end of the loop can surely be removed.
Another look at the situation is that we can pick any point $$$i$$$, and then pick point $$$j$$$ that is the closest to point $$$i$$$, either by Euclidean or Manhattan metric. Pair $$$(i, j)$$$ can be removed then, as if any point $$$k$$$ lies inside the bounding box of $$$i$$$ and $$$j$$$, it's strictly closer to $$$i$$$ than $$$j$$$.
Both of these solutions work in $$$O(n^2)$$$.