Yandex.Algorithm 2011: Finals

Solutions for Yandex.Algorithm 2011: Finals (contest 97). 4/5 problems verified against sample I/O. Difficulty range: 2200-2700.

5 items

Yandex.Algorithm 2011: Finals

Type: Special | Problems: 5 | Verified: 4/5 | Rating range: 2200-2700 | Time: 11m 24s

Problem	Name	Rating	Tags	Solve Time	Verified
A	Domino	2400	brute-force, implementation	2m 46s	✗
B	Superset	2300	constructive-algorithms, divide-and-conquer	2m 12s	✓
C	Winning Strategy	2400	binary-search, graphs, math	1m 44s	✓
D	Robot in Basement	2700	bitmasks, brute-force, implementation	2m 39s	✓
E	Leaders	2200	dfs-and-similar, dsu, graphs	2m 3s	✓

Practice › Codeforces Solutions › Yandex.Algorithm 2011: Finals ›

CF 97A - Domino

We are given a grid of size n × m, representing a board where Gennady has placed 28 domino chips. Each domino occupies exactly two adjacent squares, and the squares of the same domino are marked with the same letter, while different dominoes have different letters.

Practice › Codeforces Solutions › Yandex.Algorithm 2011: Finals ›

CF 97B - Superset

We start with a set of distinct lattice points on the plane. We may add more points, and the final set must satisfy a geometric condition for every pair of points. Take any two points.

Practice › Codeforces Solutions › Yandex.Algorithm 2011: Finals ›

CF 97C - Winning Strategy

Each year the university sends a team of exactly n students to the finals. Some of those students may already have participated once before, and the rest are newcomers.

Practice › Codeforces Solutions › Yandex.Algorithm 2011: Finals ›

CF 97D - Robot in Basement

We have a grid representing a basement. Some cells are walls, some are walkable, and exactly one walkable cell is the exit. A robot starts in an unknown walkable cell. We are given a fixed sequence of movement commands such as L, R, U, D.

Practice › Codeforces Solutions › Yandex.Algorithm 2011: Finals ›

CF 97E - Leaders

We are given an undirected graph where vertices represent people and edges represent relationships. For every query (u, v), we must decide whether there exists a simple path from u to v whose length is odd. The keyword here is "simple". We are not allowed to revisit vertices.