# --- Day 12: Passage Pathing ---

# With your submarine's subterranean subsystems subsisting suboptimally, the
# only way you're getting out of this cave anytime soon is by finding a path
# yourself. Not just a path - the only way to know if you've found the best
# path is to find all of them.

# Fortunately, the sensors are still mostly working, and so you build a rough
# map of the remaining caves (your puzzle input). For example:

# start-A
# start-b
# A-c
# A-b
# b-d
# A-end
# b-end

# This is a list of how all of the caves are connected. You start in the cave
# named start, and your destination is the cave named end. An entry like b-d
# means that cave b is connected to cave d - that is, you can move between
# them.

# So, the above cave system looks roughly like this:

#     start
#     /   \
# c--A-----b--d
#     \   /
#      end

# Your goal is to find the number of distinct paths that start at start, end at
# end, and don't visit small caves more than once. There are two types of
# caves: big caves (written in uppercase, like A) and small caves (written in
# lowercase, like b). It would be a waste of time to visit any small cave more
# than once, but big caves are large enough that it might be worth visiting
# them multiple times. So, all paths you find should visit small caves at most
# once, and can visit big caves any number of times.

# Given these rules, there are 10 paths through this example cave system:

# start,A,b,A,c,A,end
# start,A,b,A,end
# start,A,b,end
# start,A,c,A,b,A,end
# start,A,c,A,b,end
# start,A,c,A,end
# start,A,end
# start,b,A,c,A,end
# start,b,A,end
# start,b,end

# (Each line in the above list corresponds to a single path; the caves visited
# by that path are listed in the order they are visited and separated by
# commas.)

# Note that in this cave system, cave d is never visited by any path: to do so,
# cave b would need to be visited twice (once on the way to cave d and a second
# time when returning from cave d), and since cave b is small, this is not
# allowed.

# Here is a slightly larger example:

# dc-end
# HN-start
# start-kj
# dc-start
# dc-HN
# LN-dc
# HN-end
# kj-sa
# kj-HN
# kj-dc

# The 19 paths through it are as follows:

# start,HN,dc,HN,end
# start,HN,dc,HN,kj,HN,end
# start,HN,dc,end
# start,HN,dc,kj,HN,end
# start,HN,end
# start,HN,kj,HN,dc,HN,end
# start,HN,kj,HN,dc,end
# start,HN,kj,HN,end
# start,HN,kj,dc,HN,end
# start,HN,kj,dc,end
# start,dc,HN,end
# start,dc,HN,kj,HN,end
# start,dc,end
# start,dc,kj,HN,end
# start,kj,HN,dc,HN,end
# start,kj,HN,dc,end
# start,kj,HN,end
# start,kj,dc,HN,end
# start,kj,dc,end

# Finally, this even larger example has 226 paths through it:

# fs-end
# he-DX
# fs-he
# start-DX
# pj-DX
# end-zg
# zg-sl
# zg-pj
# pj-he
# RW-he
# fs-DX
# pj-RW
# zg-RW
# start-pj
# he-WI
# zg-he
# pj-fs
# start-RW

# How many paths through this cave system are there that visit small caves at
# most once?

from collections import defaultdict
from typing import DefaultDict

caves_map = defaultdict(list)
with open("files/P12.txt") as f:
    for row in f:
        start, end = row.strip().split("-")
        caves_map[start].append(end)
        caves_map[end].append(start)


def dfs(
    node: str,
    graph: DefaultDict[str, list[str]],
    visited: set[str],
    already_visited: bool,
    counter: int = 0,
):
    if node == "end":
        return 1
    for neighbour in graph[node]:
        if neighbour.isupper():
            counter += dfs(neighbour, graph, visited, already_visited)
        else:
            if neighbour not in visited:
                counter += dfs(
                    neighbour, graph, visited | {neighbour}, already_visited
                )
            elif already_visited and neighbour not in {"start", "end"}:
                counter += dfs(neighbour, graph, visited, False)
    return counter


def part_1() -> None:
    ans = dfs("start", caves_map, {"start"}, already_visited=False)
    print(f"There are {ans} paths visiting small caves")


# --- Part Two ---

# After reviewing the available paths, you realize you might have time to visit
# a single small cave twice. Specifically, big caves can be visited any number
# of times, a single small cave can be visited at most twice, and the remaining
# small caves can be visited at most once. However, the caves named start and
# end can only be visited exactly once each: once you leave the start cave, you
# may not return to it, and once you reach the end cave, the path must end
# immediately.

# Now, the 36 possible paths through the first example above are:

# start,A,b,A,b,A,c,A,end
# start,A,b,A,b,A,end
# start,A,b,A,b,end
# start,A,b,A,c,A,b,A,end
# start,A,b,A,c,A,b,end
# start,A,b,A,c,A,c,A,end
# start,A,b,A,c,A,end
# start,A,b,A,end
# start,A,b,d,b,A,c,A,end
# start,A,b,d,b,A,end
# start,A,b,d,b,end
# start,A,b,end
# start,A,c,A,b,A,b,A,end
# start,A,c,A,b,A,b,end
# start,A,c,A,b,A,c,A,end
# start,A,c,A,b,A,end
# start,A,c,A,b,d,b,A,end
# start,A,c,A,b,d,b,end
# start,A,c,A,b,end
# start,A,c,A,c,A,b,A,end
# start,A,c,A,c,A,b,end
# start,A,c,A,c,A,end
# start,A,c,A,end
# start,A,end
# start,b,A,b,A,c,A,end
# start,b,A,b,A,end
# start,b,A,b,end
# start,b,A,c,A,b,A,end
# start,b,A,c,A,b,end
# start,b,A,c,A,c,A,end
# start,b,A,c,A,end
# start,b,A,end
# start,b,d,b,A,c,A,end
# start,b,d,b,A,end
# start,b,d,b,end
# start,b,end

# The slightly larger example above now has 103 paths through it, and the even
# larger example now has 3509 paths through it.

# Given these new rules, how many paths through this cave system are there?


def part_2() -> None:
    ans = dfs("start", caves_map, {"start"}, already_visited=True)
    print(f"There are {ans} paths in total")


if __name__ == "__main__":
    part_1()
    part_2()