SansPapyrus683

use itertools::Itertools;
use std::collections::{HashMap, HashSet, VecDeque};
use std::fs;

fn neighbors(r: usize, c: usize) -> Vec<(usize, usize)> {
    let (ir, ic) = (r as i32, c as i32);
    vec![(ir + 1, ic), (ir - 1, ic), (ir, ic + 1), (ir, ic - 1)]
        .iter()
        .filter(|(r, c)| *r >= 0 && *c >= 0)
        .map(|(r, c)| (*r as usize, *c as usize))

SansPapyrus683

" https://stackoverflow.com/a/234578/12128483
filetype plugin indent on
" show existing tab with 4 spaces width
set tabstop=4
" when indenting with '>', use 4 spaces width
set shiftwidth=4
" On pressing tab, insert 4 spaces
set expandtab  " set expandtab& to just indent w/ tabs
" https://vi.stackexchange.com/a/4250
set softtabstop=4

SansPapyrus683

#include <iostream>
#include <vector>
#include <deque>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>

template <typename T>
std::ostream& operator<<(std::ostream& out, const std::vector<T>& vec) {

SansPapyrus683

import json
import re
import os
import shutil
import requests


def load_twt_obj(file: str) -> list:
    raw = open(file).read()
    return json.loads(raw[raw.find("=") + 1:])

SansPapyrus683

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using std::cout;
using std::endl;
using std::vector;
using std::string;

SansPapyrus683

Translation credit to Kaz Nakao

You are ordering $N$ matryoshka dolls. They each have a diameter $R$ and a height $H$. A matryoshka doll will only fit inside another if the former's height and width are both strictly less than the latter's.

You also have $Q$ queries. Each of these $Q$ queries has two arguments: $A$ and $B$. For each of the queries, you need to answer the following question: Among the set of dolls that have minimum diameter $A$ and maximum height $B$, how can you fit these dolls in each other, making "towers" of dolls, so that the total number of "towers" is minimal?

SansPapyrus683

Translation credit to Kaz Nakao

In the country of IOI, there was a railway mogul JOI that owned all of the railroads.
However, JOI has passed away and the railways he owns are to be distributed as a split inheritance.

In IOI, there are $N$ cities and there are $M$ railroads that connect the cities.
The cities are numbered $1$ to $N$, and the railroads are numbered $1$ to $M$.
A railroad $i$ runs between city $A_i$ and $B_i$ in both directions and generates a distinct profit of $C$ yen.
There may be several lines between the same two cities.

SansPapyrus683

bro i could passed gold lmao
i actually knew how to solve 2 problems

but anyways, here's the gold p1 editorial on request (actual problem here, i won't bother explaining it myself, just read the darn thing)

so it would be absolute cancer to try and actually calculate if each tuple can get to a level where all cows have the same hunger level
not only that, calculating all the possible tuples themselves...

SansPapyrus683

general code stuff

• comments
  • inline comments should be two spaces after the line of code
  • if a single comment takes over two lines, use a multiline comment
• sols should have nearly the same structure, same names, etc. to avoid confusion
• avoid global variables
• for the brace languages, always use braces & space em out like: for (int i = 0; i < 2; i++) {
• an if-else should be like this:

SansPapyrus683

so yknow i was on the dp grind when someone posted this problem on the github issues for the usaco guide (you can test here, but prepare google translate)

so we have a buncha circles, intervals really
and we wanna remove some so that none of them intersect (they can be tangent tho)

so let's define min_remove[s][e] as the minimum amount of circles we have to remove given that the only circles we consider are the ones that are completely