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CP Weekly Challenge
CCC 2026 J4 – Snail Path Solution & Analysis
Simulate the snail step by step, track slimy squares with a visited set, and count every revisit.
Imagine the snail starts on one grid square. That starting square is already part of its trail, because the problem says the snail makes the squares it touches slimy. In the example “move east 3 squares”, there are 4 slimy squares total, which includes the starting square.
本题在 CCC DSA 地图里的位置
- Level: Junior J4
- Pattern: Grid Simulation, Coordinate State, Visited Set
- Prerequisite: loops, while/for simulation, string parsing, coordinates, sets
- What this teaches: 把一条看似完整的移动指令拆成逐格状态更新,并在正确时机检查重复访问。
- Where it appears again: grid adjacency, flood fill, BFS, implicit graph search
下一步推荐
- 相关 CP Weekly 真题:CCC 2023 J4/S1 Trianglane(grid adjacency)、CCC 2024 J5 Harvest Waterloo(flood fill)、CCC 2020 J5/S2 Escape Room(implicit graph BFS)
- 相关 Python / C++ 基础:loops and accumulation、while simulation、strings parsing、sets、2D coordinates
- 相关 DSA 专题:CCC Grid Problems: from simulation to graph search(planned)
Where This Fits in the CCC DSA Map
- Level: Junior J4
- Pattern: Grid Simulation, Coordinate State, Visited Set
- Prerequisite: loops, while/for simulation, string parsing, coordinates, sets
- What this teaches: Split a movement command into unit steps, update state carefully, and check repeated visits at the right time.
- Where it appears again: grid adjacency, flood fill, BFS, implicit graph search
What to Study Next
- Related CP Weekly problems: CCC 2023 J4/S1 Trianglane for grid adjacency, CCC 2024 J5 Harvest Waterloo for flood fill, CCC 2020 J5/S2 Escape Room for implicit graph BFS.
- Related Python / C++ foundations: loops and accumulation, while simulation, string parsing, sets, 2D coordinates.
- Related DSA roadmap article: CCC Grid Problems: from simulation to graph search (planned).
Understanding the Problem
Every time the snail moves one square, we check whether the new square was already slimy.
- If it steps onto a square it has never touched before, nothing is added to the answer.
- If it steps onto a square that is already slimy, the snail has returned to an old square, so we add 1 to the answer.
So the problem is really asking: as the snail walks step by step, how many times does it revisit an old square?
Important Detail Students Often Miss
A movement like S2 does not mean one big jump. It means move 1 square south, then move 1 more square south.
You must process movements one square at a time, because the snail could enter an already slimy square in the middle of a longer movement. The movement format is a direction letter followed by a positive integer up to 20.
Core DSA Concepts
- Simulation: follow the snail exactly as described, updating its position and trail step by step.
- Hash set / visited set: remember which squares are already slimy using coordinate pairs such as
(x, y). - Coordinates on a grid: use
xfor east/west andyfor north/south. - Unit-step processing: break each command into single-square moves before checking whether the square was already visited.
Coordinate Updates
- North:
y += 1 - South:
y -= 1 - East:
x += 1 - West:
x -= 1
Big Idea
Start from the original square and add it to a visited set. For each movement command, repeat the movement one square at a time. After each single step, check whether the new coordinate is already in the set. If it is, increase the revisit count. Then add the coordinate to the set.
Python Solution
# CCC 2026 J4 - Snail Path
# Read the number of movements
M = int(input())
# Current position of the snail
x = 0
y = 0
# The starting square is already touched, so it is slimy
visited = set()
visited.add((x, y))
# Count how many times the snail enters a square
# that was already slimy
answer = 0
# Direction changes for one step
dx = {
"E": 1,
"W": -1,
"N": 0,
"S": 0
}
dy = {
"E": 0,
"W": 0,
"N": 1,
"S": -1
}
# Process each movement
for _ in range(M):
move = input().strip()
# First character is the direction
direction = move[0]
# The rest of the string is the distance
distance = int(move[1:])
# Move one square at a time
for _ in range(distance):
x += dx[direction]
y += dy[direction]
if (x, y) in visited:
answer += 1
visited.add((x, y))
print(answer)
Common Mistakes
- Treating
S20as one jump: that is wrong. The snail passes through 20 squares one by one, and any of those may already be slimy. - Forgetting the starting square: the starting square must already be in
visited, otherwise returning to start will not be counted correctly. - Counting distinct repeated squares instead of total re-entries: the output is how many times the snail enters a slimy square, not how many slimy squares are revisited. A single square can be entered many times and should be counted many times. The statement explicitly notes that one slimy square can be entered multiple times.
- Using a list instead of a set: a set is much faster for checking whether a coordinate has been seen before.
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