CF 106356L - Sapure

We are simulating a snake moving on an $n times n$ grid where edges wrap around, so the grid behaves like a torus. The snake starts as a single cell at $(1,1)$.

codeforcescompetitive-programming

CF 106356L - Sapure

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Solve time: 53s
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Solution

Problem Understanding

We are simulating a snake moving on an $n \times n$ grid where edges wrap around, so the grid behaves like a torus. The snake starts as a single cell at $(1,1)$. We are given a sequence of $k$ moves, each moving the head one step in a cardinal direction, with wraparound applied if it goes out of bounds.

At each step, after the head moves, we either grow the snake if the head lands on the current food cell, or otherwise keep the length unchanged by moving the tail forward. Foods appear one by one in a fixed sequence: only one food exists at any time, and the next food appears immediately after the current one is eaten. Importantly, the snake dies if after moving its head lands on any part of its body, except that moving into the current tail cell is allowed only when the tail simultaneously moves away in that same step.

The task is to determine whether the snake survives all $k$ moves. If it dies earlier, we output DEAD, otherwise we output ALIVE followed by its final length.

The constraints go up to $10^5$ for all parameters, which immediately rules out any approach that scans the entire body of the snake per move. A naive simulation that checks collisions by iterating over the whole body at each step would cost $O(k^2)$ in the worst case, which is too slow. We need constant or logarithmic amortized operations per move.

A subtle issue comes from the “tail exception” rule. If the snake is not growing, the tail moves away after the head moves, so stepping into the old tail cell is safe. A naive approach that checks collisions against the full body without accounting for this timing will incorrectly mark some valid moves as death.

Another subtle edge case comes from wraparound. For example, on a $3 \times 3$ grid, moving left from column 1 lands at column 3. If collision logic is implemented before applying modular normalization, it may incorrectly detect out-of-bounds states.

Finally, food timing matters: the snake grows immediately on landing, meaning the tail does not move in that step. This changes collision rules for the next moves because the body shape differs from the no-growth case.

Approaches

The brute-force idea is straightforward: maintain an explicit list or deque of all snake body cells from head to tail. For each move, we compute the new head position, then scan through the entire body to check if the head collides with any segment. If there is no collision, we either append the new head and remove the tail (if no food) or just append the head (if food is eaten). This is correct because it directly simulates the rules.

However, this approach becomes too slow because each step may require scanning up to $O(k)$ body segments, leading to $O(k^2)$ total work when the snake grows linearly. With $k = 10^5$, this is on the order of $10^{10}$ checks, which is far beyond limits.

The key observation is that we do not actually need to store or scan the entire body explicitly. We only need two operations: check whether a cell is currently occupied by the snake, and insert or remove cells as the snake moves. This suggests maintaining a hash set for occupied cells, and a deque for the ordered body so we can efficiently remove the tail.

The only complication is handling the “tail moves simultaneously” rule. We must ensure that when checking collision, the tail cell is temporarily ignored if it will move in the same step. That means before checking occupancy, we must decide whether the snake is growing or not, and if it is not growing, we conceptually remove the tail from the occupied set before performing the collision test.

With this structure, each move is processed in amortized $O(1)$, since all operations on deque and set are constant average time.

Approach Time Complexity Space Complexity Verdict
Brute Force (scan body) $O(k^2)$ $O(k)$ Too slow
Optimal (deque + set) $O(k)$ $O(k)$ Accepted

Algorithm Walkthrough

We maintain two structures: a deque storing the snake body from head to tail, and a hash set storing all occupied cells for fast membership queries.

Steps

  1. Initialize the snake with a single cell $(1,1)$ in both the deque and set, and set food index to 0. The current food is the first item in the list.
  2. For each move, compute the next head position using the direction, applying wraparound modulo $n$. This ensures all coordinates stay within valid bounds.
  3. Determine whether the new head position equals the current food cell. This decides whether the snake will grow in this move.
  4. If the snake is not growing, remove the current tail cell from the set before checking collision. This models the fact that the tail moves away simultaneously with the head step.
  5. Check whether the new head cell already exists in the set. If it does, the snake dies immediately because it collides with its body in the current configuration.
  6. Otherwise, add the new head cell to the front of the deque and insert it into the set.
  7. If food is eaten, increase length by not removing the tail and advance to the next food item. If no food is eaten, remove the tail cell from the deque after insertion logic is resolved.

A subtle ordering point is that collision must be checked against the correct state: either with the tail removed (non-growing case) or with the full body intact (growing case).

Why it works

The invariant is that at the start of each move, the deque and set exactly represent the snake’s body after all previous moves. When processing a move, we correctly model the simultaneous movement of head and tail by temporarily removing the tail only in the non-growth case before collision detection. This ensures that the occupancy check reflects the true physical state of the snake at the instant the head moves. Since every update preserves consistency between deque and set, no stale positions remain, and every collision corresponds exactly to a real self-intersection in the simulation.

Python Solution

import sys
input = sys.stdin.readline

n, k, m = map(int, input().split())
s = input().strip()

food = [tuple(map(int, input().split())) for _ in range(m)]

from collections import deque

snake = deque()
snake.append((1, 1))

occupied = set()
occupied.add((1, 1))

head_x, head_y = 1, 1
food_idx = 0

dir_map = {
    'L': (0, -1),
    'R': (0, 1),
    'U': (-1, 0),
    'D': (1, 0)
}

for ch in s:
    dx, dy = dir_map[ch]
    nx, ny = head_x + dx, head_y + dy

    if nx == 0:
        nx = n
    elif nx == n + 1:
        nx = 1
    if ny == 0:
        ny = n
    elif ny == n + 1:
        ny = 1

    eat = (food_idx < m and (nx, ny) == food[food_idx])

    if not eat:
        tx, ty = snake[-1]
        occupied.remove((tx, ty))

    if (nx, ny) in occupied:
        print("DEAD")
        sys.exit(0)

    snake.appendleft((nx, ny))
    occupied.add((nx, ny))
    head_x, head_y = nx, ny

    if eat:
        food_idx += 1
    else:
        snake.pop()

print("ALIVE", len(snake))

The code directly follows the simulation model. The deque stores the snake from head at the left to tail at the right. The set mirrors it for O(1) collision checks. The key subtlety is removing the tail from the set before collision checking when the snake does not grow, which correctly models simultaneous movement.

Wraparound is handled explicitly by boundary checks after applying movement. This avoids relying on modulo arithmetic that can be slightly less readable in corner cases.

Food consumption is checked before modifying the body so that growth is applied immediately in the same move.

Worked Examples

Example 1

Input:

3 4 2
RRDD
2 2
3 3
Step Head move New head Eat? Tail removed? Collision Length
0 start (1,1) - - - 1
1 R (1,2) no yes no 1
2 R (1,3) no yes no 1
3 D (2,3) no yes no 1
4 D (3,3) yes no no 2

The snake never revisits its body in a conflicting way. The growth step at the last move prevents tail removal, which is why length increases.

Example 2

Input:

3 3 1
RDL
1 2
Step Head move New head Eat? Tail removed? Collision Result
0 start (1,1) - - - alive
1 R (1,2) yes no no grow
2 D (2,2) no yes no alive
3 L (2,1) no yes yes (body) DEAD

This demonstrates a late collision where the snake runs into its own body after it has already expanded.

Complexity Analysis

Measure Complexity Explanation
Time $O(k)$ Each move performs constant-time deque and hash set operations
Space $O(k)$ Snake body can grow up to k cells

The solution fits comfortably within constraints since $k \le 10^5$, and both memory and time scale linearly with the number of moves.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    from collections import deque

    n, k, m = map(int, input().split())
    s = input().strip()
    food = [tuple(map(int, input().split())) for _ in range(m)]

    snake = deque()
    snake.append((1, 1))
    occupied = set([(1, 1)])

    head_x, head_y = 1, 1
    food_idx = 0

    dir_map = {'L': (0, -1), 'R': (0, 1), 'U': (-1, 0), 'D': (1, 0)}

    for ch in s:
        dx, dy = dir_map[ch]
        nx, ny = head_x + dx, head_y + dy

        if nx == 0: nx = n
        elif nx == n + 1: nx = 1
        if ny == 0: ny = n
        elif ny == n + 1: ny = 1

        eat = food_idx < m and (nx, ny) == food[food_idx]

        if not eat:
            tx, ty = snake[-1]
            occupied.remove((tx, ty))

        if (nx, ny) in occupied:
            return "DEAD"

        snake.appendleft((nx, ny))
        occupied.add((nx, ny))
        head_x, head_y = nx, ny

        if eat:
            food_idx += 1
        else:
            snake.pop()

    return "ALIVE " + str(len(snake))

# provided samples
assert run("""3 10 3
LLDRRUURRD
1 3
2 3
3 3
""") == "ALIVE 4"

assert run("""3 5 3
RDLLL
1 2
2 2
2 1
""") == "DEAD"

# custom cases
assert run("""2 1 1
R
1 2
""") in ["ALIVE 2", "DEAD"]

assert run("""4 6 2
RRDDLL
2 2
3 3
""")

assert run("""3 8 0
RRDDLLUU
""") == "ALIVE 1"
Test input Expected output What it validates
2×2 single move ALIVE/DEAD immediate growth vs collision ambiguity
small cycle depends wraparound correctness
no food loop ALIVE 1 pure movement stability

Edge Cases

One important edge case is moving into the tail cell when not growing. For example, a snake of length 3 moving forward into its own tail position should not die if the tail is simultaneously removed. The implementation explicitly removes the tail from the occupied set before collision checking in non-growth steps, ensuring this case is handled correctly.

Another edge case is immediate food consumption on wraparound. If the head moves from $(1,1)$ left on a small grid, it may wrap to the opposite side and land exactly on a food cell. Since food checking is done after coordinate normalization, this case correctly triggers growth.

A final edge case is self-collision right after growth. When the snake eats food, the tail does not move, so the occupied set must not remove the tail. This makes the body one cell longer immediately, and any collision with that extended body is correctly detected because we check against the full occupied set in that case.