CF 102697121 - Space Invaders

Rating: -
Tags: -
Solve time: 1m
Verified: yes

Solution

Problem Understanding

The task is a direct three-dimensional geometry calculation. The spaceship has one fixed position (X, Y, Z), and the radar reports n alien positions. For every alien, we need to compute its Euclidean distance from the spaceship and print the distances in the same order as the aliens were given.

For an alien at (x, y, z), the three coordinate differences are x - X, y - Y, and z - Z. Squaring them removes their signs, adding the three squared differences gives the squared distance, and taking the square root gives the actual distance:

distance = sqrt((x-X)^2 + (y-Y)^2 + (z-Z)^2).

The official statement specifies a 1 second time limit and 256 MB of memory. It does not expose a numeric upper bound for the coordinates or for n on the current problem page, so the safest interpretation is to design the solution to scale linearly with the number of aliens. Since the output itself contains one value for every alien, an algorithm that spends constant work per alien is asymptotically optimal. A quadratic algorithm would become unusable as n grows, while an O(n) solution only performs one fixed-size calculation per input point. Python integers also avoid overflow in the intermediate squared differences, although the normal coordinate bounds would make overflow a non-issue in languages with sufficiently wide integer types.

There are several small cases where an implementation can silently go wrong. When the alien is exactly at the spaceship, the answer must be zero. For example,

1 2 3
1
1 2 3

produces

0.0

A careless implementation that forgets the square root would print 0, which is numerically equal but does not follow the required decimal output format.

Negative coordinate differences must also be squared correctly. For example,

2 4 6
1
-2 -4 -6

produces

14.966629547095765

Taking absolute values before squaring happens to give the same result, but using only the raw sum of differences would incorrectly allow positive and negative coordinates to cancel.

The order of the aliens matters. For example,

produces

3.0
4.0

An implementation that sorts the aliens before processing them would change the required output order.

Approaches

The most direct approach is already the optimal one. For every alien, compute the three coordinate differences, square them, add the three values, take the square root, and print the result. This is correct because the formula is exactly the Euclidean distance definition in three dimensions, so each alien can be handled independently of every other alien.

If there are n aliens, each alien requires three subtractions, three multiplications, two additions, and one square-root operation. Ignoring input and output costs, that is nine fixed arithmetic operations per alien, or 9n such operations for the entire input. The exact machine-level cost of a square root depends on the implementation, but it is still constant with respect to n.

A more elaborate brute-force interpretation might try to compare every alien with every other point, search for geometric relationships, or repeatedly recompute distances. None of that is useful here because the requested distance depends only on the spaceship and the current alien. There is no interaction between aliens that needs to be discovered.

The key observation is that the problem is completely separable. Once the spaceship coordinates are known, the answer for alien i can be calculated without knowing anything about alien j. That means there is no dynamic programming state, graph traversal, sorting, or geometric data structure to maintain. We simply perform the constant-size formula once per alien.

The output requirement also gives a lower bound of Omega(n), because the program has to produce n separate answers. Consequently, O(n) time is asymptotically optimal. The optimal solution is therefore the same simple computation as the brute-force baseline, with the crucial difference that no unnecessary pairwise work is introduced.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n) when implemented as one distance calculation per alien	O(1)	Accepted
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

Read the spaceship coordinates X, Y, and Z. These coordinates remain unchanged for every distance calculation.
Read n, the number of detected aliens. We need exactly one output value for each of these n positions.
For each alien, read its coordinates (x, y, z). Compute dx = x - X, dy = y - Y, and dz = z - Z. Keeping the differences signed is fine because the next operation squares them.
Compute dx * dx + dy * dy + dz * dz. This is the squared Euclidean distance, so it avoids any geometric approximation before the final square root.
Take the square root of that sum and print it immediately. There is no need to store previous aliens or their distances because every answer is independent.
Repeat the calculation for all n aliens, preserving their input order in the output.

Why it works

For every alien, the algorithm calculates exactly the three coordinate differences between the alien and the spaceship. Squaring those differences and adding them produces the squared Euclidean distance, and taking the square root produces the Euclidean distance itself. Since the calculation is performed independently for every alien and the aliens are processed in input order, every printed value is the required distance in the required position. No information from another alien can affect the result, so processing the points independently cannot miss a condition or produce a different valid interpretation.

Python Solution

import sys
import math

input = sys.stdin.readline

def solve():
    X, Y, Z = map(int, input().split())
    n = int(input())

    out = []

    for _ in range(n):
        x, y, z = map(int, input().split())

        dx = x - X
        dy = y - Y
        dz = z - Z

        distance = math.sqrt(dx * dx + dy * dy + dz * dz)
        out.append(str(distance))

    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    solve()

The first input line is stored once because the spaceship never moves. The next line gives the number of iterations needed.

For each alien, the code computes the three differences separately. This makes the formula easy to verify and avoids accidentally mixing a spaceship coordinate with an alien coordinate.

The expression inside math.sqrt is an integer squared distance. Python's arbitrary-precision integers make the intermediate calculation safe even if the coordinates are larger than the range of a fixed-width C++ int. math.sqrt then converts the non-negative value to a floating-point distance.

The answers are accumulated in out and written together. This avoids making a separate system call for every alien, which is preferable when n is large. The calculation itself uses only a constant amount of working memory apart from the output strings.

There is no rounding code because the statement explicitly says that rounding is unnecessary. Python's floating-point conversion produces the usual decimal representation expected by the judge.

Worked Examples

Sample 1

The spaceship is at (7, 3, 8). The algorithm processes six aliens independently.

Alien	dx	dy	dz	Squared distance	Distance
`(2, 1, 5)`	-5	-2	-3	38	6.164414002968976
`(3, 7, 6)`	-4	4	-2	36	6.0
`(4, 2, 1)`	-3	-1	-7	59	7.681145747868608
`(9, 8, 4)`	2	5	-4	45	6.708203932499369
`(7, 3, 8)`	0	0	0	0	0.0
`(5, 11, 16)`	-2	8	8	132	11.489125293076057

The fifth alien is exactly at the spaceship, so all three differences are zero and the algorithm naturally produces 0.0. This demonstrates that no special case is needed for coincident points.

Sample 2

The spaceship is at (2, 4, 6), and there are three aliens.

Alien	dx	dy	dz	Squared distance	Distance
`(-2, -4, -6)`	-4	-8	-12	224	14.966629547095765
`(2, -4, -6)`	0	-8	-12	208	14.422205101855956
`(-2, 4, 6)`	-4	0	0	16	4.0

The example contains negative coordinates and zero differences. It demonstrates why each coordinate must be handled independently and squared before the three components are combined.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Every alien requires a constant number of arithmetic operations and one square root.
Space	O(n)	The implementation stores the `n` output strings before writing them.

The algorithm performs only one fixed-size calculation for each detected alien, so its running time grows linearly with the input size. Since the output itself has n lines, linear time is asymptotically optimal. The mathematical working state is O(1), and the implementation uses O(n) memory only because it batches the output. If memory were extremely constrained, each answer could instead be printed immediately, reducing auxiliary memory to O(1).

Test Cases

import sys
import io
import math

def solve():
    input = sys.stdin.readline

    X, Y, Z = map(int, input().split())
    n = int(input())

    out = []

    for _ in range(n):
        x, y, z = map(int, input().split())

        dx = x - X
        dy = y - Y
        dz = z - Z

        out.append(str(math.sqrt(dx * dx + dy * dy + dz * dz)))

    sys.stdout.write("\n".join(out))

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample 1
assert run(
    """7 3 8
6
2 1 5
3 7 6
4 2 1
9 8 4
7 3 8
5 11 16
"""
) == (
    """6.164414002968976
6.0
7.681145747868608
6.708203932499369
0.0
11.489125293076057"""
), "sample 1"

# Provided sample 2
assert run(
    """2 4 6
3
-2 -4 -6
2 -4 -6
-2 4 6
"""
) == (
    """14.966629547095765
14.422205101855956
4.0"""
), "sample 2"

# Minimum-size input, alien is at the spaceship.
assert run(
    """0 0 0
1
0 0 0
"""
) == "0.0", "minimum size and zero distance"

# One-axis distance with a negative coordinate.
assert run(
    """10 20 30
1
-5 20 30
"""
) == "15.0", "single-axis negative difference"

# All aliens are identical and must produce identical answers.
assert run(
    """1 1 1
4
4 5 1
4 5 1
4 5 1
4 5 1
"""
) == "5.0\n5.0\n5.0\n5.0", "repeated identical aliens"

# Large stress-sized case. Every point is at distance 5 from the origin.
large_n = 100000
large_input = "0 0 0\n" + str(large_n) + "\n" + ("3 4 0\n" * large_n)
large_output = run(large_input)

assert large_output.count("\n") == large_n - 1, "large output line count"
assert all(line == "5.0" for line in large_output.splitlines()), "large repeated case"

Test input	Expected output	What it validates
`0 0 0 / 1 / 0 0 0`	`0.0`	Minimum-size input and coincident spaceship and alien
`10 20 30 / 1 / -5 20 30`	`15.0`	Negative coordinate difference and movement along only one axis
`1 1 1 / 4 / 4 5 1` repeated	Four lines containing `5.0`	Repeated equal positions and preservation of output order
`0 0 0 / 100000 / 3 4 0` repeated	`100000` lines containing `5.0`	Large input size, repeated calculations, and output handling

The current statement page does not expose a numeric maximum for n, so the final test uses 100000 as a stress-sized input rather than claiming it is the official maximum. The algorithm remains linear regardless of the exact hidden upper bound.

Edge Cases

When the alien and spaceship occupy the same point, the input

1 2 3
1
1 2 3

gives dx = 0, dy = 0, and dz = 0. The squared distance is 0, so math.sqrt(0) produces 0.0. No special branch is required, and the output is exactly

0.0

Negative coordinates are handled directly by subtraction and squaring. For

2 4 6
1
-2 -4 -6

the differences are -4, -8, and -12. Their squares are 16, 64, and 144, giving a squared distance of 224. The square root is 14.966629547095765, so the algorithm produces

14.966629547095765

A zero difference in one or two dimensions is also naturally handled. For

10 20 30
1
-5 20 30

the differences are -15, 0, and 0. The squared distance is 225, giving

15.0

This catches implementations that accidentally omit a coordinate or mishandle zero differences.

Finally, every alien must be processed independently while preserving input order. With

the three distances are 3.0, 4.0, and 5.0, so the output is

3.0
4.0
5.0

The algorithm prints each result in the same iteration in which its alien is read, so there is no sorting or indexing step that could reorder the answers.