Letter Exchange
Contest: VK Cup 2022 - Финальный раунд (Engine)
Difficulty rating: 1900
Tags: constructive algorithms

A cooperative game is played by $$$m$$$ people. In the game, there are $$$3m$$$ sheets of paper: $$$m$$$ sheets with letter 'w', $$$m$$$ sheets with letter 'i', and $$$m$$$ sheets with letter 'n'.

Initially, each person is given three sheets (possibly with equal letters).

The goal of the game is to allow each of the $$$m$$$ people to spell the word "win" using their sheets of paper. In other words, everyone should have one sheet with letter 'w', one sheet with letter 'i', and one sheet with letter 'n'.

To achieve the goal, people can make exchanges. Two people participate in each exchange. Both of them choose exactly one sheet of paper from the three sheets they own and exchange it with each other.

Find the shortest sequence of exchanges after which everyone has one 'w', one 'i', and one 'n'.

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$m$$$ ($$$2 \le m \le 10^5$$$) — the number of people.

The $$$i$$$-th of the next $$$m$$$ lines contains a string $$$s_i$$$ of length $$$3$$$ consisting of lowercase English letters 'w', 'i', and 'n', denoting the letters person $$$i$$$ has on their sheets of paper at the beginning of the game, in arbitrary order.

Each of the letters 'w', 'i', and 'n' appears on the sheets of paper exactly $$$m$$$ times in total.

It is guaranteed that the sum of $$$m$$$ over all test cases does not exceed $$$10^5$$$.

Output format:
For each test case, print a non-negative integer $$$k$$$ — the smallest number of exchanges people need to make everyone have one 'w', one 'i', and one 'n'.

In each of the next $$$k$$$ lines print four tokens $$$a_1$$$ $$$c_1$$$ $$$a_2$$$ $$$c_2$$$, describing the exchanges in chronological order ($$$1 \le a_1, a_2 \le m$$$; $$$a_1 \ne a_2$$$; $$$c_1, c_2$$$ are one of $$$\{\mathtt{w}, \mathtt{i}, \mathtt{n}\}$$$): person $$$a_1$$$ gives letter $$$c_1$$$ to person $$$a_2$$$, while person $$$a_2$$$ gives letter $$$c_2$$$ to person $$$a_1$$$ at the same time.

If there are multiple solutions, print any.

Constraints:
Time limit: 4.0 s
Memory limit: 512.0 MB

Example input:
3
2
nwi
inw
3
inn
nww
wii
4
win
www
iii
nnn

Example output:
0
2
2 w 3 i
3 w 1 n
3
2 w 3 i
2 w 4 n
3 i 4 n

Editorial:
For each person, there are three essential cases of what they could initially have:
1. Three distinct letters: "win". No need to take part in any exchanges.
2. Two equal letters and another letter, e.g. "wii". An extra 'i' must be exchanged with someone's 'n'.
3. Three equal letters, e.g. "www". One 'w' must be exchanged with someone's 'i', another 'w' must be exchanged with someone's 'n'.
Let's create a graph on three vertices: 'w', 'i', 'n'. Whenever person $$$i$$$ has an extra letter $$$x$$$ and is lacking letter $$$y$$$, create a directed edge $$$x \rightarrow y$$$ marked with $$$i$$$.
Once the graph is built, whenever you have a cycle of length $$$2$$$, that is, $$$x \xrightarrow{i} y \xrightarrow{j} x$$$, it means person $$$i$$$ needs to exchange $$$x$$$ for $$$y$$$, while person $$$j$$$ needs to exchange $$$y$$$ for $$$x$$$. Thus, both of their needs can be satisfied with just one exchange.
Finally, once there are no cycles of length $$$2$$$, note that the in-degree and the out-degree of every vertex are equal. If e.g. there are $$$p$$$ edges 'w' $$$\rightarrow$$$ 'i', it follows that there are $$$p$$$ edges 'i' $$$\rightarrow$$$ 'n' and $$$p$$$ edges 'n' $$$\rightarrow$$$ 'w'. It means we can form $$$p$$$ cycles of length $$$3$$$. (The cycles could also go in the opposite direction: 'w' $$$\rightarrow$$$ 'n' $$$\rightarrow$$$ 'i' $$$\rightarrow$$$ 'w'.) In any case, each cycle of length $$$3$$$ can be solved using $$$2$$$ exchanges.