MITIT Winter 2025 Advanced Round 2

5 problems from MITIT Winter 2025 Advanced Round 2 (contest 105667), difficulty -. 5/5 solutions verified against sample I/O.

5 items

MITIT Winter 2025 Advanced Round 2

Special | 5 problems | 5/5 verified | Difficulty - | 4m 9s

#	Problem	Time	✓
A	Toy Marbles	49s	✓
B	Snakes on a Grid	51s	✓
C	MIT Tour	55s	✓
D	Path Partition	47s	✓
E	Colored Blocks	47s	✓

Practice › Codeforces Solutions › MITIT Winter 2025 Advanced Round 2 ›

CF 105667E - Colored Blocks

We are given a sequence of colored blocks, represented as an array where each position contains a color identifier. The task is to split this sequence into the minimum number of subsequences such that each subsequence satisfies a monotonic consistency condition on colors.

Practice › Codeforces Solutions › MITIT Winter 2025 Advanced Round 2 ›

CF 105667D - Path Partition

We are given a graph where the task is to decompose its edges into simple paths, each of length exactly three edges. In other words, every edge must belong to exactly one path, and every path must consist of four distinct vertices connected consecutively.

Practice › Codeforces Solutions › MITIT Winter 2025 Advanced Round 2 ›

CF 105667C - MIT Tour

The structure described in this problem is a rooted tree where each vertex represents a room and each edge represents a corridor with a travel cost. From the root, every vertex has a well-defined depth equal to its distance in edges from the root.

Practice › Codeforces Solutions › MITIT Winter 2025 Advanced Round 2 ›

CF 105667B - Snakes on a Grid

We are working on a grid where each cell is either filled or empty, and each query gives us a rectangular subgrid. For every query, we need to decide whether the filled cells inside that rectangle form a valid structure called a snake.

Practice › Codeforces Solutions › MITIT Winter 2025 Advanced Round 2 ›

CF 105667A - Toy Marbles

We are given a system of $N$ containers, each initially holding a single marble. Every container has a target container where its marble is supposed to end up. That target is given by an array $c$, where $ci$ tells us the destination of the marble currently in container $i$.