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internbootcamp/bootcamp/ebearanddrawing/ebearanddrawing.py Executable file
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"""#
### 谜题描述
Limak is a little bear who learns to draw. People usually start with houses, fences and flowers but why would bears do it? Limak lives in the forest and he decides to draw a tree.
Recall that tree is a connected graph consisting of n vertices and n - 1 edges.
Limak chose a tree with n vertices. He has infinite strip of paper with two parallel rows of dots. Little bear wants to assign vertices of a tree to some n distinct dots on a paper so that edges would intersect only at their endpoints — drawn tree must be planar. Below you can see one of correct drawings for the first sample test.
<image>
Is it possible for Limak to draw chosen tree?
Input
The first line contains single integer n (1 ≤ n ≤ 105).
Next n - 1 lines contain description of a tree. i-th of them contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) denoting an edge between vertices ai and bi. It's guaranteed that given description forms a tree.
Output
Print \"Yes\" (without the quotes) if Limak can draw chosen tree. Otherwise, print \"No\" (without the quotes).
Examples
Input
8
1 2
1 3
1 6
6 4
6 7
6 5
7 8
Output
Yes
Input
13
1 2
1 3
1 4
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13
Output
No
Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
```python
#include <bits/stdc++.h>
using namespace std;
int n;
vector<vector<int> > adj;
vector<bool> top;
vector<int> branched;
vector<bool> visited;
vector<bool> fertile;
int root = -1;
void fail() {
cout << \"No\" << endl;
exit(0);
}
void dfsFindRoot(int u) {
if (root != -1) return;
visited[u] = true;
bool leaf = true;
for (int child : adj[u]) {
if (visited[child]) continue;
leaf = false;
}
if (!leaf) {
branched[u] = 0;
for (int child : adj[u]) {
if (visited[child]) continue;
dfsFindRoot(child);
branched[u] += branched[child];
}
if (root != -1) return;
if (branched[u] >= 3) {
root = u;
}
}
}
void dfs(int u) {
visited[u] = true;
bool leaf = true;
for (int child : adj[u]) {
if (visited[child]) continue;
leaf = false;
}
if (!leaf) {
branched[u] = 0;
int numTops = 0;
for (int child : adj[u]) {
if (visited[child]) continue;
dfs(child);
if (fertile[child]) top[u] = true;
branched[u] += branched[child];
if (top[child]) {
numTops++;
top[u] = true;
}
}
if (u != root && numTops > 1)
fail();
else if (numTops > 2)
fail();
if (branched[u] >= 3) {
top[u] = true;
} else if (branched[u] == 2)
fertile[u] = true;
}
}
int main() {
int m;
cin >> n;
adj = vector<vector<int> >(n, vector<int>(0));
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
u--;
v--;
adj[u].push_back(v);
adj[v].push_back(u);
}
if (n <= 4) {
cout << \"YES\" << endl;
return 0;
}
top = vector<bool>(n, false);
branched = vector<int>(n, 1);
visited = vector<bool>(n, false);
dfsFindRoot(0);
if (root == -1) {
cout << \"Yes\" << endl;
return 0;
}
top = vector<bool>(n, false);
branched = vector<int>(n, 1);
visited = vector<bool>(n, false);
fertile = vector<bool>(n, false);
dfs(root);
cout << \"Yes\" << endl;
}
```
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。
"""
from bootcamp import Basebootcamp
from bootcamp import Basebootcamp
import random
import re
class Ebearanddrawingbootcamp(Basebootcamp):
def __init__(self, **params):
"""
初始化谜题参数，默认n的范围为5-20
"""
self.params = params
self.params.setdefault('min_n', 5)
self.params.setdefault('max_n', 20)
def case_generator(self):
"""
生成随机树结构并计算正确答案
"""
n = random.randint(self.params['min_n'], self.params['max_n'])
edges = self._generate_random_tree(n)
correct_answer = self.is_tree_planar(n, edges)
return {
'n': n,
'edges': edges,
'correct_answer': correct_answer
}
@staticmethod
def _generate_random_tree(n):
"""
使用Prüfer序列生成随机树
"""
if n == 1:
return []
if n == 2:
return [(1, 2)]
prufer = [random.randint(1, n) for _ in range(n-2)]
degree = [1] * (n + 1)
for node in prufer:
degree[node] += 1
edges = []
for node in prufer:
for v in range(1, n+1):
if degree[v] == 1:
edges.append((node, v))
degree[node] -= 1
degree[v] -= 1
break
u = [i for i in range(1, n+1) if degree[i] == 1]
edges.append((u[0], u[1]))
return edges
@staticmethod
def is_tree_planar(n, edges):
"""
判断树是否可以被平面绘制
"""
if n <= 4:
return "Yes"
adj = [[] for _ in range(n)]
for u, v in edges:
adj[u-1].append(v-1)
adj[v-1].append(u-1)
root = -1
visited = [False] * n
branched = [1] * n
def dfs_find_root(u):
nonlocal root
if root != -1:
return
visited[u] = True
has_child = False
for child in adj[u]:
if not visited[child]:
has_child = True
dfs_find_root(child)
branched[u] += branched[child]
if has_child and branched[u] >= 3:
root = u
dfs_find_root(0)
if root == -1:
return "Yes"
top = [False] * n
branched = [1] * n
visited = [False] * n
fertile = [False] * n
failed = False
def dfs(u):
nonlocal failed
if failed:
return
visited[u] = True
leaf = True
for child in adj[u]:
if not visited[child]:
leaf = False
if not leaf:
branched[u] = 0
num_tops = 0
for child in adj[u]:
if visited[child]:
continue
dfs(child)
if fertile[child]:
top[u] = True
branched[u] += branched[child]
if top[child]:
num_tops += 1
top[u] = True
if (u != root and num_tops > 1) or (u == root and num_tops > 2):
failed = True
if branched[u] >= 3:
top[u] = True
elif branched[u] == 2:
fertile[u] = True
visited = [False] * n
dfs(root)
return "Yes" if not failed else "No"
@staticmethod
def prompt_func(question_case):
"""
生成格式化的谜题问题
"""
n = question_case['n']
edges = '\n'.join(f"{u} {v}" for u, v in question_case['edges'])
problem = f"""Limak wants to draw a tree on a special paper strip with two rows of dots. The tree must be drawn with no edge crossings except at vertices. Determine if this is possible for the given tree structure.
Input:
{n}
{edges}
Output your answer exactly as "[answer]Yes[/answer]" or "[answer]No[/answer]" after analyzing the structure."""
return problem
@staticmethod
def extract_output(output):
"""
从模型输出中提取最后一个[answer]标签内容
"""
matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL | re.IGNORECASE)
if not matches:
return None
answer = matches[-1].strip().lower()
return answer.capitalize() if answer in {'yes', 'no'} else None
@classmethod
def _verify_correction(cls, solution, identity):
"""
验证答案是否正确
"""
return solution == identity['correct_answer']