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161 lines
4.3 KiB
Python
Executable file
161 lines
4.3 KiB
Python
Executable file
"""#
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### 谜题描述
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On a plane are n points (xi, yi) with integer coordinates between 0 and 106. The distance between the two points with numbers a and b is said to be the following value: <image> (the distance calculated by such formula is called Manhattan distance).
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We call a hamiltonian path to be some permutation pi of numbers from 1 to n. We say that the length of this path is value <image>.
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Find some hamiltonian path with a length of no more than 25 × 108. Note that you do not have to minimize the path length.
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Input
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The first line contains integer n (1 ≤ n ≤ 106).
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The i + 1-th line contains the coordinates of the i-th point: xi and yi (0 ≤ xi, yi ≤ 106).
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It is guaranteed that no two points coincide.
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Output
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Print the permutation of numbers pi from 1 to n — the sought Hamiltonian path. The permutation must meet the inequality <image>.
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If there are multiple possible answers, print any of them.
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It is guaranteed that the answer exists.
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Examples
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Input
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5
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0 7
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8 10
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3 4
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5 0
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9 12
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Output
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4 3 1 2 5
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Note
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In the sample test the total distance is:
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<image>
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(|5 - 3| + |0 - 4|) + (|3 - 0| + |4 - 7|) + (|0 - 8| + |7 - 10|) + (|8 - 9| + |10 - 12|) = 2 + 4 + 3 + 3 + 8 + 3 + 1 + 2 = 26
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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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#include <bits/stdc++.h>
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using namespace std;
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const long long MAXN = 1e6 + 1, SQ = 1200;
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int n;
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int x[MAXN], y[MAXN], a[MAXN];
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bool cmp(int i, int j) {
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short int tmp = x[i] / SQ, tmpp = x[j] / SQ;
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if (tmp ^ tmpp)
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return x[i] < x[j];
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else {
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if (tmp & 1)
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return y[i] > y[j];
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else
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return y[i] < y[j];
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}
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}
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int main() {
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ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
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cin >> n;
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for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
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for (int i = 0; i < n; i++) a[i] = i;
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sort(a, a + n, cmp);
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for (int i = 0; i < n; i++) cout << a[i] + 1 << ' ';
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return 0;
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}
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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 random
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import re
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from bootcamp import Basebootcamp
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class Epointsonplanebootcamp(Basebootcamp):
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def __init__(self, n=5, SQ=1200):
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self.n = n
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self.SQ = SQ
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def case_generator(self):
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n = self.n
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SQ = self.SQ
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points_set = set()
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while len(points_set) < n:
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x = random.randint(0, 10**6)
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y = random.randint(0, 10**6)
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points_set.add((x, y))
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points = [list(p) for p in points_set]
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return {
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'n': n,
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'points': points,
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'SQ': SQ
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}
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@staticmethod
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def prompt_func(question_case):
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input_lines = [str(question_case['n'])]
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for point in question_case['points']:
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input_lines.append(f"{point[0]} {point[1]}")
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input_example = '\n'.join(input_lines)
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prompt = f"""你是一个算法竞赛专家，需要解决一个哈密顿路径问题。题目要求找到一条路径，总曼哈顿距离不超过25×10^8。
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输入格式：
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- 首行为点数n（1 ≤ n ≤ 1e6)
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- 接下来n行每行两个整数，表示点的x和y坐标
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输出格式：
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- 一个1到n的排列，用空格分隔
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当前问题输入：
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{input_example}
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请将答案置于[answer]和[/answer]之间。"""
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return prompt
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not matches:
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return None
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last_answer = matches[-1].strip()
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parts = last_answer.split()
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try:
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solution = list(map(int, parts))
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return solution
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except ValueError:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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n = identity['n']
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points = identity['points']
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# 验证排列有效性
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if len(solution) != n or sorted(solution) != list(range(1, n+1)):
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return False
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# 实际计算总距离
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total = 0
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for i in range(n-1):
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idx1 = solution[i] - 1
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idx2 = solution[i+1] - 1
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x1, y1 = points[idx1]
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x2, y2 = points[idx2]
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total += abs(x1 - x2) + abs(y1 - y2)
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if total > 25e8:
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return False
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return total <= 25e8
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