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206 lines
6.8 KiB
Python
Executable file
206 lines
6.8 KiB
Python
Executable file
"""#
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### 谜题描述
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You are given n points on the plane. The polygon formed from all the n points is strictly convex, that is, the polygon is convex, and there are no three collinear points (i.e. lying in the same straight line). The points are numbered from 1 to n, in clockwise order.
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We define the distance between two points p_1 = (x_1, y_1) and p_2 = (x_2, y_2) as their Manhattan distance: $$$d(p_1, p_2) = |x_1 - x_2| + |y_1 - y_2|.$$$
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Furthermore, we define the perimeter of a polygon, as the sum of Manhattan distances between all adjacent pairs of points on it; if the points on the polygon are ordered as p_1, p_2, …, p_k (k ≥ 3), then the perimeter of the polygon is d(p_1, p_2) + d(p_2, p_3) + … + d(p_k, p_1).
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For some parameter k, let's consider all the polygons that can be formed from the given set of points, having any k vertices, such that the polygon is not self-intersecting. For each such polygon, let's consider its perimeter. Over all such perimeters, we define f(k) to be the maximal perimeter.
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Please note, when checking whether a polygon is self-intersecting, that the edges of a polygon are still drawn as straight lines. For instance, in the following pictures:
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<image>
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In the middle polygon, the order of points (p_1, p_3, p_2, p_4) is not valid, since it is a self-intersecting polygon. The right polygon (whose edges resemble the Manhattan distance) has the same order and is not self-intersecting, but we consider edges as straight lines. The correct way to draw this polygon is (p_1, p_2, p_3, p_4), which is the left polygon.
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Your task is to compute f(3), f(4), …, f(n). In other words, find the maximum possible perimeter for each possible number of points (i.e. 3 to n).
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Input
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The first line contains a single integer n (3 ≤ n ≤ 3⋅ 10^5) — the number of points.
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Each of the next n lines contains two integers x_i and y_i (-10^8 ≤ x_i, y_i ≤ 10^8) — the coordinates of point p_i.
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The set of points is guaranteed to be convex, all points are distinct, the points are ordered in clockwise order, and there will be no three collinear points.
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Output
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For each i (3≤ i≤ n), output f(i).
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Examples
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Input
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4
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2 4
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4 3
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3 0
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1 3
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Output
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12 14
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Input
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3
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0 0
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0 2
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2 0
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Output
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8
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Note
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In the first example, for f(3), we consider four possible polygons:
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* (p_1, p_2, p_3), with perimeter 12.
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* (p_1, p_2, p_4), with perimeter 8.
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* (p_1, p_3, p_4), with perimeter 12.
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* (p_2, p_3, p_4), with perimeter 12.
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For f(4), there is only one option, taking all the given points. Its perimeter 14.
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In the second example, there is only one possible polygon. Its perimeter is 8.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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n = int(raw_input())
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X = []
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Y = []
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for _ in range(n):
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x,y = map(int,raw_input().strip(' ').strip('\n').split(' '))
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X.append(x)
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Y.append(y)
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maxx,minx = max(X),min(X)
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maxy,miny = max(Y),min(Y)
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ans = -10e18
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for i in range(n):
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dx = max(maxx-X[i],X[i]-minx)
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dy = max(maxy-Y[i],Y[i]-miny)
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ans = max(ans,dx+dy)
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ans = str(2*ans)+' '
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rec = 2*(maxx+maxy-minx-miny)
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for i in range(3,n):
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ans += str(rec)+' '
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ans = ans.strip('')
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print ans
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import re
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import random
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from bootcamp import Basebootcamp
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class Coptimalpolygonperimeterbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.min_points = params.get('min_points', 3)
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self.max_points = params.get('max_points', 10)
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self.coord_range = params.get('coord_range', (-100, 100))
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# Ensure min_points is at least 3
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self.min_points = max(3, self.min_points)
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def case_generator(self):
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# Generate n in specified range
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n = random.randint(self.min_points, self.max_points)
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# Generate bounding box with sufficient size
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minx = random.randint(self.coord_range[0], self.coord_range[1] - 10)
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maxx = minx + 10 # Ensure width is at least 10
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miny = random.randint(self.coord_range[0], self.coord_range[1] - 10)
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maxy = miny + 10 # Ensure height is at least 10
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# Generate points in strict convex polygon (simulating convex hull)
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points = []
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edges = ['bottom', 'right', 'top', 'left']
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edge_idx = 0
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# Generate points in clockwise order without colinear points
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for _ in range(n):
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edge = edges[edge_idx % 4]
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edge_idx += 1
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if edge == 'bottom':
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x = random.randint(minx, maxx - 1) # Avoid rightmost point
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y = miny
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elif edge == 'right':
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x = maxx
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y = random.randint(miny + 1, maxy - 1) # Avoid top/bottom extremes
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elif edge == 'top':
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x = random.randint(minx + 1, maxx) # Avoid leftmost point
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y = maxy
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else: # left
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x = minx
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y = random.randint(miny + 1, maxy - 1)
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points.append((x, y))
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# Calculate expected output using problem logic
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X = [p[0] for p in points]
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Y = [p[1] for p in points]
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maxx_val = max(X)
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minx_val = min(X)
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maxy_val = max(Y)
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miny_val = min(Y)
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ans_3 = 0
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for x, y in zip(X, Y):
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dx = max(maxx_val - x, x - minx_val)
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dy = max(maxy_val - y, y - miny_val)
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ans_3 = max(ans_3, dx + dy)
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ans_3 *= 2
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rec = 2 * ((maxx_val - minx_val) + (maxy_val - miny_val))
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# Generate expected output sequence
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expected_output = [ans_3] + [rec] * (n - 3)
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return {
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'n': n,
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'points': points,
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'expected_output': expected_output
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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input_lines = [str(question_case['n'])] + [
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f"{x} {y}" for x, y in question_case['points']
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]
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input_example = '\n'.join(input_lines)
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return f"""\
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Given a strictly convex polygon with {question_case['n']} vertices arranged clockwise. Compute maximum perimeter for k=3..n using Manhattan distances.
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Input:
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{input_example}
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Output format: Space-separated integers f(3) to f(n).
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Place your answer within [answer][/answer] tags. Example: [answer]12 14[/answer]"""
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@staticmethod
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def extract_output(output):
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answer_blocks = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not answer_blocks:
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return None
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return answer_blocks[-1].strip()
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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solution_values = list(map(int, solution.split()))
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return solution_values == identity['expected_output']
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except:
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return False
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