202 lines
6.2 KiB
Rust
202 lines
6.2 KiB
Rust
use std::collections::{HashMap, VecDeque};
|
|
|
|
type Name = (char, char);
|
|
|
|
#[derive(Debug)]
|
|
struct Valve {
|
|
name: (char, char),
|
|
flow_rate: i32,
|
|
tunnels: Vec<Name>,
|
|
}
|
|
|
|
impl Valve {
|
|
fn print(&self) {
|
|
print!("{}{} -- {:3} -- ", self.name.0, self.name.1, self.flow_rate);
|
|
for t in &self.tunnels {
|
|
print!("{}{}, ", t.0, t.1);
|
|
}
|
|
println!();
|
|
}
|
|
}
|
|
|
|
fn main() {
|
|
// Use command line arguments to specify the input filename.
|
|
let args: Vec<String> = std::env::args().collect();
|
|
if args.len() < 2 {
|
|
panic!("Usage: ./main <input-file>\nNo input file provided. Exiting.");
|
|
}
|
|
|
|
// Next, read the contents of the input file into a string for easier processing.
|
|
let input = std::fs::read_to_string(&args[1]).expect("Error opening file");
|
|
|
|
// --- TASK BEGIN ---
|
|
|
|
// Parse the input.
|
|
// Collect all nodes by name to a big map.
|
|
let mut nodes: HashMap<Name, Valve> = HashMap::new();
|
|
for line in input.lines() {
|
|
// Split word-by-word.
|
|
let words = line.split_whitespace().collect::<Vec<_>>();
|
|
|
|
// Determine this node's name.
|
|
let name = (
|
|
words[1].chars().next().unwrap(),
|
|
words[1].chars().nth(1).unwrap(),
|
|
);
|
|
|
|
// Grab the list of outgoing nodes for this node and strip the whitespace.
|
|
// Returns "DD,II,BB".
|
|
let tunnel_nodes = line
|
|
.split("valve")
|
|
.nth(1)
|
|
.unwrap()
|
|
.chars()
|
|
.filter(|e| *e != ' ' && *e != 's')
|
|
.collect::<String>();
|
|
// Next, split by ','.
|
|
// Returns ["DD", "II", "BB"].
|
|
let tunnel_nodes = tunnel_nodes.split(',').collect::<Vec<_>>();
|
|
// Turn the vector into a vector of names.
|
|
// Returns [('D', 'D'), ('I', 'I'), ('B', 'B')].
|
|
let tunnel_nodes: Vec<Name> = tunnel_nodes
|
|
.iter()
|
|
.map(|e| (e.chars().next().unwrap(), e.chars().nth(1).unwrap()))
|
|
.collect();
|
|
|
|
// Construct this node.
|
|
let node = Valve {
|
|
name,
|
|
flow_rate: words[4]
|
|
.strip_prefix("rate=")
|
|
.unwrap()
|
|
.strip_suffix(';')
|
|
.unwrap()
|
|
.parse()
|
|
.unwrap(),
|
|
tunnels: tunnel_nodes,
|
|
};
|
|
// Add this node to the big map.
|
|
nodes.insert(name, node);
|
|
}
|
|
|
|
// In order to properly calculate the optimal path and valve order
|
|
// we need to first compute the cost getting from any node A to any
|
|
// other node B, i.e. perform pathfinding.
|
|
// We will precompute the results for faster lookup times later.
|
|
let mut distances: HashMap<(Name, Name), i32> = HashMap::new();
|
|
for an in nodes.keys() {
|
|
// Technically we're performing Dijkstra's shortest path algorithm.
|
|
// Since the edges all have weight 1 this devolves to simple breadth-first search.
|
|
|
|
// Keep track of all nodes in a queue.
|
|
let mut q: VecDeque<(Name, i32)> = VecDeque::new();
|
|
// Add the current node to that queue.
|
|
q.push_back((*an, 0));
|
|
|
|
loop {
|
|
// Queue empty? Then we're done.
|
|
if q.is_empty() {
|
|
break;
|
|
}
|
|
|
|
// Grab the next element (n) and its distance (d) from the queue.
|
|
let (n, d) = q.pop_front().unwrap();
|
|
|
|
// Since this is Dijkstra's algorithm, nodes are only visited once.
|
|
// Therefore, add this node to the proper result.
|
|
distances.insert((*an, n), d);
|
|
|
|
// Next, look at all of this node's neighbors and add them to the queue.
|
|
for neighbor in &nodes[&n].tunnels {
|
|
// Only add them if they haven't already been visited.
|
|
if distances.get(&(*an, *neighbor)).is_none() {
|
|
q.push_back((*neighbor, d + 1));
|
|
}
|
|
}
|
|
}
|
|
}
|
|
|
|
// Keep track of all nodes with non-zero flow_rate.
|
|
let mut non_zero_nodes: Vec<Name> = nodes
|
|
.iter()
|
|
.filter(|(_, v)| v.flow_rate > 0)
|
|
.map(|(k, _)| *k)
|
|
.collect();
|
|
|
|
// Now check through all possible permutations of non-zero nodes using a recursive function.
|
|
// We assume 'AA' has zero flow_rate (it should).
|
|
let mut optimum: i32 = 0;
|
|
generate_permutation(
|
|
&nodes,
|
|
&distances,
|
|
&mut non_zero_nodes,
|
|
&mut vec![('A', 'A')],
|
|
30,
|
|
0,
|
|
&mut optimum,
|
|
);
|
|
|
|
println!("Optimal pressure release: {}", optimum);
|
|
}
|
|
|
|
fn generate_permutation(
|
|
nodes: &HashMap<Name, Valve>,
|
|
distances: &HashMap<(Name, Name), i32>,
|
|
source: &mut Vec<Name>,
|
|
dest: &mut Vec<Name>,
|
|
time: i32,
|
|
pressure: i32,
|
|
optimum: &mut i32,
|
|
) {
|
|
// Are we done here?
|
|
// If there are only 2 units of time (or less) left we're done here.
|
|
// We're obviously also done if the source is empty.
|
|
if time <= 2 || source.is_empty() {
|
|
// Check if this is better and store the optimal result.
|
|
if pressure > *optimum {
|
|
*optimum = pressure;
|
|
}
|
|
// Obviously, return early.
|
|
return;
|
|
}
|
|
// Iterate through all elements remaining in the source.
|
|
for i in 0..source.len() {
|
|
// Remove the current element from the vector.
|
|
let e = source.remove(i);
|
|
|
|
// Determine the distance between the last two nodes.
|
|
let add_dist = distances[&(*dest.last().unwrap(), e)];
|
|
|
|
// Put the element onto the destination.
|
|
dest.push(e);
|
|
|
|
// Determine how much time will have passed until this node is ready.
|
|
let new_time = time - (1 + add_dist);
|
|
// Calculate how much pressure we save.
|
|
let add_pressure = new_time * nodes[&e].flow_rate;
|
|
|
|
// Finally, call recursively.
|
|
generate_permutation(
|
|
nodes,
|
|
distances,
|
|
source,
|
|
dest,
|
|
new_time,
|
|
pressure + add_pressure,
|
|
optimum,
|
|
);
|
|
|
|
// Remove the element from the destination.
|
|
dest.pop();
|
|
// And reinsert the element back into the vector, at the same precise location.
|
|
source.insert(i, e);
|
|
}
|
|
}
|
|
|
|
fn print_name_list(list: &Vec<Name>) {
|
|
for n in list {
|
|
print!("{}{}, ", n.0, n.1);
|
|
}
|
|
println!();
|
|
}
|