Overview

This problem asks you to select a base from \(N\) settlements and, for each of \(Q\) queries, compute the sum of Euclidean distances (each truncated) from that base to all settlements.

Analysis

Naive Approach

For each query, computing the distance from the base to all \(N\) settlements requires \(O(N)\) computation per query. Computing all \(Q\) queries naively results in \(O(NQ)\) total time.

Since \(N \leq 2000\) and \(Q \leq 2 \times 10^5\), \(NQ\) can be up to \(4 \times 10^8\), which risks TLE if implemented directly.

Key Insight: Deduplication of Queries

The base must be placed at one of the settlements (\(1\) through \(N\)). Therefore, there are at most \(N\) distinct base candidates. Even if the same settlement is specified in multiple queries, the answer is the same.

By computing only for unique queries (base candidates) and caching the results, the computational complexity is reduced to \(O(N^2)\) (at most \(N\) types of bases × distance calculation to each of \(N\) settlements).

Since \(N \leq 2000\), \(N^2 = 4 \times 10^6\), which is sufficiently fast.

Algorithm

1. Read the coordinates of all settlements.
2. Read all queries and determine the set of unique base candidates.
3. For each unique base candidate \(c\), compute the sum of distances to all settlements \(j\): $\(\text{results}[c] = \sum_{j=1}^{N} \lfloor \sqrt{(X_c - X_j)^2 + (Y_c - Y_j)^2} \rfloor\)\( By using integer square root (`isqrt`), we can compute \)\lfloor \sqrt{d^2} \rfloor$ exactly without floating-point errors.
4. For each query, output the precomputed result.

Concrete Example

For example, if there are 3 settlements at \((0, 0)\), \((3, 4)\), \((1, 1)\), and the base is placed at settlement 1 (coordinates \((0,0)\)): - Distance to settlement 1: \(\lfloor\sqrt{0}\rfloor = 0\) - Distance to settlement 2: \(\lfloor\sqrt{9+16}\rfloor = \lfloor 5 \rfloor = 5\) - Distance to settlement 3: \(\lfloor\sqrt{1+1}\rfloor = \lfloor\sqrt{2}\rfloor = 1\)

The total is \(0 + 5 + 1 = 6\).

Complexity

• Time complexity: \(O(N^2 + Q)\)
  • There are at most \(N\) unique base candidates, and for each we compute distances to \(N\) settlements, giving \(O(N^2)\)
  • Output for queries takes \(O(Q)\)
• Space complexity: \(O(N + Q)\)
  • \(O(N)\) for settlement coordinates, \(O(N + Q)\) for storing queries and caching results

Implementation Notes

• Using isqrt: Using floating-point math.sqrt may yield incorrect floor values due to rounding errors. Python’s math.isqrt returns the exact truncated integer square root, allowing us to compute \(\lfloor\sqrt{d^2}\rfloor\) without error.

• Query deduplication: By using a set to compute only unique queries, the computation stays at \(O(N^2)\) even when \(Q\) is large.

• Fast I/O: By reading all input at once with sys.stdin.buffer.read() and writing output collectively with '\n'.join(), TLE can be more easily avoided even in Python.

  Source Code

import sys
import math

def main():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    Q = int(input_data[idx]); idx += 1
    
    X = [0] * N
    Y = [0] * N
    for i in range(N):
        X[i] = int(input_data[idx]); idx += 1
        Y[i] = int(input_data[idx]); idx += 1
    
    # Precompute answers for each unique base
    # N, Q <= 2000, so O(N*N) = 4*10^6 which is fine
    # But queries may repeat, so we cache by village index
    
    from math import isqrt
    
    results = {}
    
    queries = []
    for k in range(Q):
        c = int(input_data[idx]) - 1; idx += 1
        queries.append(c)
    
    unique_queries = set(queries)
    
    for c in unique_queries:
        cx, cy = X[c], Y[c]
        total = 0
        for j in range(N):
            dx = cx - X[j]
            dy = cy - Y[j]
            d2 = dx * dx + dy * dy
            total += isqrt(d2)
        results[c] = total
    
    out = []
    for c in queries:
        out.append(str(results[c]))
    
    sys.stdout.write('\n'.join(out) + '\n')

main()

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