Jeremy Bearimy
Contest: Codeforces Round 607 (Div. 1)
Difficulty rating: 2000
Tags: dfs and similar, graphs, greedy, trees

Welcome! Everything is fine.

You have arrived in The Medium Place, the place between The Good Place and The Bad Place. You are assigned a task that will either make people happier or torture them for eternity.

You have a list of $$$k$$$ pairs of people who have arrived in a new inhabited neighborhood. You need to assign each of the $$$2k$$$ people into one of the $$$2k$$$ houses. Each person will be the resident of exactly one house, and each house will have exactly one resident.

Of course, in the neighborhood, it is possible to visit friends. There are $$$2k - 1$$$ roads, each of which connects two houses. It takes some time to traverse a road. We will specify the amount of time it takes in the input. The neighborhood is designed in such a way that from anyone's house, there is exactly one sequence of distinct roads you can take to any other house. In other words, the graph with the houses as vertices and the roads as edges is a tree.

The truth is, these $$$k$$$ pairs of people are actually soulmates. We index them from $$$1$$$ to $$$k$$$. We denote by $$$f(i)$$$ the amount of time it takes for the $$$i$$$-th pair of soulmates to go to each other's houses.

As we have said before, you will need to assign each of the $$$2k$$$ people into one of the $$$2k$$$ houses. You have two missions, one from the entities in The Good Place and one from the entities of The Bad Place. Here they are:

- The first mission, from The Good Place, is to assign the people into the houses such that the sum of $$$f(i)$$$ over all pairs $$$i$$$ is minimized. Let's define this minimized sum as $$$G$$$. This makes sure that soulmates can easily and efficiently visit each other;
- The second mission, from The Bad Place, is to assign the people into the houses such that the sum of $$$f(i)$$$ over all pairs $$$i$$$ is maximized. Let's define this maximized sum as $$$B$$$. This makes sure that soulmates will have a difficult time to visit each other.

What are the values of $$$G$$$ and $$$B$$$?

Input format:
The first line of input contains a single integer $$$t$$$ ($$$1 \le t \le 500$$$) denoting the number of test cases. The next lines contain descriptions of the test cases.

The first line of each test case contains a single integer $$$k$$$ denoting the number of pairs of people ($$$1 \le k \le 10^5$$$). The next $$$2k - 1$$$ lines describe the roads; the $$$i$$$-th of them contains three space-separated integers $$$a_i, b_i, t_i$$$ which means that the $$$i$$$-th road connects the $$$a_i$$$-th and $$$b_i$$$-th houses with a road that takes $$$t_i$$$ units of time to traverse ($$$1 \le a_i, b_i \le 2k$$$, $$$a_i \neq b_i$$$, $$$1 \le t_i \le 10^6$$$). It is guaranteed that the given roads define a tree structure.

It is guaranteed that the sum of the $$$k$$$ in a single file is at most $$$3 \cdot 10^5$$$.

Output format:
For each test case, output a single line containing two space-separated integers $$$G$$$ and $$$B$$$.

Constraints:
Time limit: 3.0 s
Memory limit: 256.0 MB

Note:
For the sample test case, we have a minimum sum equal to $$$G = 15$$$. One way this can be achieved is with the following assignment:

- The first pair of people get assigned to houses $$$5$$$ and $$$6$$$, giving us $$$f(1) = 5$$$;
- The second pair of people get assigned to houses $$$1$$$ and $$$4$$$, giving us $$$f(2) = 6$$$;
- The third pair of people get assigned to houses $$$3$$$ and $$$2$$$, giving us $$$f(3) = 4$$$.

Note that the sum of the $$$f(i)$$$ is $$$5 + 6 + 4 = 15$$$.

We also have a maximum sum equal to $$$B = 33$$$. One way this can be achieved is with the following assignment:

- The first pair of people get assigned to houses $$$1$$$ and $$$4$$$, giving us $$$f(1) = 6$$$;
- The second pair of people get assigned to houses $$$6$$$ and $$$2$$$, giving us $$$f(2) = 14$$$;
- The third pair of people get assigned to houses $$$3$$$ and $$$5$$$, giving us $$$f(3) = 13$$$.

Note that the sum of the $$$f(i)$$$ is $$$6 + 14 + 13 = 33$$$.

Example input:
2
3
1 2 3
3 2 4
2 4 3
4 5 6
5 6 5
2
1 2 1
1 3 2
1 4 3

Example output:
15 33
6 6

Editorial:
Maximization
Suppose we're maximizing the sum. Consider a single edge $$$(a,b)$$$, and consider the two components on either side of this edge. Then we have an important observation: in the optimal solution, the nodes of one component are all paired with nodes on the other component. This is because otherwise, there will be at least one pair in each component that lies entirely in that component, say $$$(i_a, j_a)$$$ and $$$(i_b, j_b$$$). But if we switch the pairing to, say, $$$(i_a, i_b)$$$ and $$$(j_a, j_b)$$$, then the cost increases, because we're introducing new edges (namely $$$(a,b)$$$, among possibly others) while keeping everything from the previous pairing.
Repeating this, we can construct an optimal solution where all nodes in one component are paired with nodes in the other component.
This means that in the optimal solution, the edge $$$(a, b)$$$ is counted exactly $$$\min(c_a, c_b)$$$ times, where $$$c_a$$$ is the size of the component on $$$a$$$'s side, and $$$c_b$$$ is the size of the component on $$$b$$$'s side. Therefore, the edge contributes exactly $$$\mathit{weight}(a,b) \cdot \min(c_a, c_b)$$$ to the answer.
But the same is true for all edges! Therefore, we can compute the answer by just summing up all contributions. The only remaining step needed is to compute the sizes of all subtrees, and this can be done with a single BFS/DFS and DP. This runs in $$$O(n)$$$.
Minimization
Now, suppose we're minimizing the sum. Consider again a single edge $$$(a,b)$$$. Again, we have an important observation: in the optimal solution, at most one pair passes through $$$(a,b)$$$. This is because otherwise, if there are at least two such pairs, then we can again switch the pairing (essentially the reverse of maximizing), which decreases the cost, because it doesn't introduce additional edges but it decreases the number of pairs passing through $$$(a,b)$$$ by $$$2$$$.
Repeating this, we can ensure that at most one pair passes through $$$(a,b)$$$.
Furthermore, the parity of the number of pairs passing through $$$(a,b)$$$ is fixed. (Why?) Therefore, in the optimal solution, $$$(a,b)$$$ is counted exactly $$$(c_a \bmod 2)$$$ times. (Note that $$$c_a \equiv c_b \pmod{2}$$$)
But again, the same is true for all edges! Therefore, we can compute the answer in $$$O(n)$$$ as well, by summing up all contributions.