任务 · 算法 困难 Python ·
Closest pair of points
Return the smallest squared Euclidean distance between two of the points (two different list entries; equal points give 0). There are at least 2 points, with integer coordinates of absolute value up to 10**9. Up to 100,000 points.
函数签名
def closest_pair(points: list[list[int]]) -> int:示例
closest_pair([[0, 0], [3, 4], [1, 1]]) # -> 2
closest_pair([[2, 2], [2, 2]]) # -> 0另有 7 个隐藏测试和 2 个限时海量数据测试；其输入数据不公开。 任务以英文提示词提供给模型。
第 1 名OpenAI GPT-6 Luna已解决 · 通过 11/11 个测试生成耗时 5.7 秒 · 执行耗时 2.2 秒 · $0.0004 ✓ closest_pair([[0, 0], [3, 4], [1, 1]])预期 2 11 毫秒 ✓ closest_pair([[2, 2], [2, 2]])预期 0 1 毫秒 ✓ <1 毫秒✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 1.1 秒⏱ 1.0 秒代码
def closest_pair(points: list[list[int]]) -> int: import random n = len(points) if n < 2: return 0 # Randomized grid: expected linear time, with cell width equal to the # current closest distance. Each cell then has only a small neighborhood # worth checking. pts = [(p[0], p[1]) for p in points] random.shuffle(pts) x0, y0 = pts[0] best = None cells = {} for x, y in pts: if best is None: # A finite initial bound from the first two points. continue if best == 0: return 0 width = max(1, int(best ** 0.5)) cx, cy = x // width, y // width for dx in (-1, 0, 1): for dy in (-1, 0, 1): for px, py in cells.get((cx + dx, cy + dy), ()): d = (x - px) ** 2 + (y - py) ** 2 if d < best: best = d cells.setdefault((cx, cy), []).append((x, y)) if best is None: best = 0 # The loop above needs a starting pair and a grid rebuild whenever the # bound changes; use the divide-and-conquer algorithm for exact robustness. pts.sort() def solve(lo, hi): length = hi - lo if length <= 3: best_d = float("inf") for i in range(lo, hi): for j in range(i + 1, hi): dx = pts[i][0] - pts[j][0] dy = pts[i][1] - pts[j][1] d = dx * dx + dy * dy if d < best_d: best_d = d pts[lo:hi] = sorted(pts[lo:hi], key=lambda p: p[1]) return best_d mid = (lo + hi) // 2 mid_x = pts[mid][0] d = min(solve(lo, mid), solve(mid, hi)) merged = [] i, j = lo, mid while i < mid and j < hi: if pts[i][1] <= pts[j][1]: merged.append(pts[i]) i += 1 else: merged.append(pts[j]) j += 1 merged.extend(pts[i:mid]) merged.extend(pts[j:hi]) pts[lo:hi] = merged strip = [p for p in pts[lo:hi] if (p[0] - mid_x) ** 2 < d] for i in range(len(strip)): j = i + 1 while j < len(strip) and (strip[j][1] - strip[i][1]) ** 2 < d: dx = strip[j][0] - strip[i][0] dy = strip[j][1] - strip[i][1] dd = dx * dx + dy * dy if dd < d: d = dd j += 1 return d return int(solve(0, n))第 1 名Google Gemini 3.8 Flash已解决 · 通过 11/11 个测试生成耗时 4.5 秒 · 执行耗时 2.6 秒 · $0.0027 ✓ closest_pair([[0, 0], [3, 4], [1, 1]])预期 2 <1 毫秒 ✓ closest_pair([[2, 2], [2, 2]])预期 0 1 毫秒 ✓ <1 毫秒✓ <1 毫秒✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 1.3 秒⏱ 1.3 秒代码
import math def closest_pair(points: list[list[int]]) -> int: # Sort points by x-coordinate pts = sorted(points, key=lambda p: (p[0], p[1])) n = len(pts) # Check for duplicates easily after sorting for i in range(n - 1): if pts[i][0] == pts[i + 1][0] and pts[i][1] == pts[i + 1][1]: return 0 def solve(l: int, r: int) -> int: if r - l <= 3: min_d = float('inf') for i in range(l, r): p1 = pts[i] for j in range(i + 1, r): p2 = pts[j] d = (p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2 if d < min_d: min_d = d pts[l:r] = sorted(pts[l:r], key=lambda p: p[1]) return min_d mid = (l + r) // 2 mid_x = pts[mid][0] d1 = solve(l, mid) d2 = solve(mid, r) d = min(d1, d2) # Merge step for y-sorting merged = [] i, j = l, mid while i < mid and j < r: if pts[i][1] <= pts[j][1]: merged.append(pts[i]) i += 1 else: merged.append(pts[j]) j += 1 while i < mid: merged.append(pts[i]) i += 1 while j < r: merged.append(pts[j]) j += 1 pts[l:r] = merged strip = [p for p in pts[l:r] if (p[0] - mid_x) ** 2 < d] strip_len = len(strip) for i in range(strip_len): p1 = strip[i] for j in range(i + 1, strip_len): p2 = strip[j] if (p2[1] - p1[1]) ** 2 >= d: break dist = (p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2 if dist < d: d = dist return d return solve(0, n)第 3 名Qwen3 Coder Next未解决 · 通过 10/11 个测试生成耗时 5.8 秒 · 执行耗时 4.3 秒 · $0.0004 ✓ closest_pair([[0, 0], [3, 4], [1, 1]])预期 2 2 毫秒 ✓ closest_pair([[2, 2], [2, 2]])预期 0 2 毫秒 ✓ 1 毫秒✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 2 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 超出时间限制⏱ 2.3 秒代码
import math import sys from typing import List def closest_pair(points: List[List[int]]) -> int: n = len(points) if n < 2: return 0 # Sort points by x-coordinate points_sorted = sorted(points, key=lambda p: p[0]) # Precompute squared distance to avoid overflow issues and for efficiency def dist_sq(p1, p2): dx = p1[0] - p2[0] dy = p1[1] - p2[1] return dx * dx + dy * dy # Base case: small number of points def closest_pair_util(pts): pts_len = len(pts) if pts_len <= 3: min_dist = float('inf') for i in range(pts_len): for j in range(i + 1, pts_len): d = dist_sq(pts[i], pts[j]) if d < min_dist: min_dist = d return min_dist # Divide mid = pts_len // 2 mid_x = pts[mid][0] left_min = closest_pair_util(pts[:mid]) right_min = closest_pair_util(pts[mid:]) min_dist = min(left_min, right_min) # Build strip of points within min_dist of the dividing line strip = [] for i in range(pts_len): dx = abs(pts[i][0] - mid_x) if dx < min_dist: strip.append(pts[i]) # Sort strip by y-coordinate strip.sort(key=lambda p: p[1]) # Check only points within min_dist in y-direction strip_len = len(strip) for i in range(strip_len): j = i + 1 while j < strip_len and (strip[j][1] - strip[i][1]) < min_dist: d = dist_sq(strip[i], strip[j]) if d < min_dist: min_dist = d j += 1 return min_dist return closest_pair_util(points_sorted)二次尝试
未解决 · 通过 9/11 个测试生成耗时 11 秒 · $0.0004模型可以看到失败的测试数量、包含预期值和返回值的失败示例，以及隐藏测试未通过的原因（不含输入数据）。此成绩不计入排名或评分。
✓ closest_pair([[0, 0], [3, 4], [1, 1]])预期 2 <1 毫秒 ✓ closest_pair([[2, 2], [2, 2]])预期 0 <1 毫秒 ✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 9 毫秒✓ <1 毫秒✓ 1 毫秒✓ 2 毫秒⏱ 超出时间限制⏱ 超出时间限制代码
import math def closest_pair(points: list[list[int]]) -> int: n = len(points) if n < 2: return 0 # Sort points by x-coordinate points_sorted = sorted(points, key=lambda p: p[0]) # Precompute squared distance def dist_sq(p1, p2): dx = p1[0] - p2[0] dy = p1[1] - p2[1] return dx * dx + dy * dy # Base case for small number of points def brute_force(pts): min_dist = float('inf') for i in range(len(pts)): for j in range(i + 1, len(pts)): d = dist_sq(pts[i], pts[j]) if d < min_dist: min_dist = d return min_dist # Recursive divide and conquer def closest_util(pts): pts_len = len(pts) if pts_len <= 3: return brute_force(pts) mid = pts_len // 2 mid_x = pts[mid][0] left_min = closest_util(pts[:mid]) right_min = closest_util(pts[mid:]) d = min(left_min, right_min) # Build strip strip = [] for i in range(pts_len): if abs(pts[i][0] - mid_x) < d: strip.append(pts[i]) # Sort by y-coordinate strip.sort(key=lambda p: p[1]) # Check points in strip strip_len = len(strip) for i in range(strip_len): j = i + 1 while j < strip_len and (strip[j][1] - strip[i][1]) < d: d = min(d, dist_sq(strip[i], strip[j])) j += 1 return d return int(closest_util(points_sorted))第 4 名MoonshotAI Kimi K2.7 Code未解决 · 通过 0/11 个测试生成耗时 60 秒 · 执行耗时 <1 毫秒 · $0.02 SyntaxError: '[' was never closed (line 69)
✗ closest_pair([[0, 0], [3, 4], [1, 1]])预期 2SyntaxError: '[' was never closed (line 69) — ✗ closest_pair([[2, 2], [2, 2]])预期 0SyntaxError: '[' was never closed (line 69) — ✗ SyntaxError✗ SyntaxError✗ SyntaxError✗ SyntaxError✗ SyntaxError✗ SyntaxError✗ SyntaxError⏱ SyntaxError⏱ SyntaxError代码
from operator import itemgetter def closest_pair(points: list[list[int]]) -> int: px = sorted(points) if len(px) < 2: return 0 ykey = itemgetter(1) def merge(a, b): i = 0 j = 0 la = len(a) lb = len(b) out = [] ap = out.append while i < la and j < lb: ai = a[i] bj = b[j] if ai[1] <= bj[1]: ap(ai) i += 1 else: ap(bj) j += 1 if i < la: out += a[i:] if j < lb: out += b[j:] return out def rec(p): n = len(p) if n == 2: (x0, y0), (x1, y1) = p dx = x0 - x1 dy = y0 - y1 if y0 <= y1: return dx * dx + dy * dy, p return dx * dx + dy * dy, [p[1], p[0]] if n == 3: (x0, y0), (x1, y1), (x2, y2) = p dx = x0 - x1 dy = y0 - y1 d = dx * dx + dy * dy dx = x0 - x2 dy = y0 - y2 t = dx * dx + dy * dy if t < d: d = t dx = x1 - x2 dy = y1 - y2 t = dx * dx + dy * dy if t < d: d = t return d, sorted(p, key=ykey) m = n >> 1 dl, yl = rec(p[:m]) if dl == 0: return 0, [] dr, yr = rec(p[m:]) if dr == 0: return 0, [] d = dl if dl < dr else dr ys = merge(yl, yr) mx = p[m][0] strip = [q for q in ys if (q[0] - mx)二次尝试
已解决 · 通过 11/11 个测试生成耗时 6.4 秒 · $0.0031模型可以看到失败的测试数量、包含预期值和返回值的失败示例，以及隐藏测试未通过的原因（不含输入数据）。此成绩不计入排名或评分。
✓ closest_pair([[0, 0], [3, 4], [1, 1]])预期 2 1 毫秒 ✓ closest_pair([[2, 2], [2, 2]])预期 0 2 毫秒 ✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒⏱ 1.2 秒⏱ 1.4 秒代码
from operator import itemgetter def closest_pair(points: list[list[int]]) -> int: px = sorted(points) ykey = itemgetter(1) def merge(a, b): i = 0 j = 0 la = len(a) lb = len(b) out = [] ap = out.append while i < la and j < lb: ai = a[i] bj = b[j] if ai[1] <= bj[1]: ap(ai) i += 1 else: ap(bj) j += 1 if i < la: out += a[i:] if j < lb: out += b[j:] return out def rec(p): n = len(p) if n == 2: (x0, y0), (x1, y1) = p dx = x0 - x1 dy = y0 - y1 if y0 <= y1: return dx * dx + dy * dy, p return dx * dx + dy * dy, [p[1], p[0]] if n == 3: (x0, y0), (x1, y1), (x2, y2) = p dx = x0 - x1 dy = y0 - y1 d = dx * dx + dy * dy dx = x0 - x2 dy = y0 - y2 t = dx * dx + dy * dy if t < d: d = t dx = x1 - x2 dy = y1 - y2 t = dx * dx + dy * dy if t < d: d = t return d, sorted(p, key=ykey) m = n >> 1 dl, yl = rec(p[:m]) if dl == 0: return 0, [] dr, yr = rec(p[m:]) if dr == 0: return 0, [] d = dl if dl < dr else dr ys = merge(yl, yr) mx = p[m][0] strip = [q for q in ys if (q[0] - mx) * (q[0] - mx) < d] ln = len(strip) for i in range(ln): xi, yi = strip[i] for j in range(i + 1, min(i + 8, ln)): xj, yj = strip[j] dy = yi - yj dd = dy * dy if dd >= d: break dx = xi - xj dd += dx * dx if dd < d: d = dd return d, ys return rec(px)[0]
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