CPMagical CP Library

Problem solution · Python

ABC273 D — LRUD Instructions

ABC273 D — LRUD Instructions: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Sorting and greedy selection
Source
KATO-Hiro AtCoder Solutions
Length
57 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sorting and greedy selection

For ABC273 D — LRUD Instructions, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 57 lines of Python from the credited upstream file abc273_d.py.
  • The implementation visibly relies on hash lookup, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from KATO-Hiro AtCoder Solutions by KATO-Hiro and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeABC273 D — LRUD Instructions · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    from collections import defaultdict    from bisect import bisect_left    import sys     input = sys.stdin.readline     h, w, y, x = map(int, input().split())    n = int(input())    # グリッドの端部のマスを用意    rows = defaultdict(lambda: [0, w + 1])    cols = defaultdict(lambda: [0, h + 1])     # 壁の位置を行と列でそれぞれ管理    for i in range(n):        ri, ci = map(int, input().split())        rows[ri].append(ci)        cols[ci].append(ri)        for key, value in rows.items():        rows[key] = sorted(value)     for key, value in cols.items():        cols[key] = sorted(value)     q = int(input())    ans = [(0, 0)] * q     for i in range(q):        di, li = input().rstrip().split()        li = int(li)         if di == "L":            index = bisect_left(rows[y], x)            x = max(x - li, rows[y][index - 1] + 1)        elif di == "R":            index = bisect_left(rows[y], x)            x = min(x + li, rows[y][index] - 1)        elif di == "U":            index = bisect_left(cols[x], y)            y = max(y - li, cols[x][index - 1] + 1)        elif di == "D":            index = bisect_left(cols[x], y)            y = min(y + li, cols[x][index] - 1)                ans[i] = (y, x)     for ri, ci in ans:        print(ri, ci)  if __name__ == "__main__":    main()

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗