mirror of
https://github.com/InternLM/InternBootcamp.git
synced 2026-04-19 12:58:04 +00:00
285 lines
7.9 KiB
Python
Executable file
285 lines
7.9 KiB
Python
Executable file
"""#
|
||
|
||
### 谜题描述
|
||
A way to make a new task is to make it nondeterministic or probabilistic. For example, the hard task of Topcoder SRM 595, Constellation, is the probabilistic version of a convex hull.
|
||
|
||
Let's try to make a new task. Firstly we will use the following task. There are n people, sort them by their name. It is just an ordinary sorting problem, but we can make it more interesting by adding nondeterministic element. There are n people, each person will use either his/her first name or last name as a handle. Can the lexicographical order of the handles be exactly equal to the given permutation p?
|
||
|
||
More formally, if we denote the handle of the i-th person as hi, then the following condition must hold: <image>.
|
||
|
||
Input
|
||
|
||
The first line contains an integer n (1 ≤ n ≤ 105) — the number of people.
|
||
|
||
The next n lines each contains two strings. The i-th line contains strings fi and si (1 ≤ |fi|, |si| ≤ 50) — the first name and last name of the i-th person. Each string consists only of lowercase English letters. All of the given 2n strings will be distinct.
|
||
|
||
The next line contains n distinct integers: p1, p2, ..., pn (1 ≤ pi ≤ n).
|
||
|
||
Output
|
||
|
||
If it is possible, output \"YES\", otherwise output \"NO\".
|
||
|
||
Examples
|
||
|
||
Input
|
||
|
||
3
|
||
gennady korotkevich
|
||
petr mitrichev
|
||
gaoyuan chen
|
||
1 2 3
|
||
|
||
|
||
Output
|
||
|
||
NO
|
||
|
||
|
||
Input
|
||
|
||
3
|
||
gennady korotkevich
|
||
petr mitrichev
|
||
gaoyuan chen
|
||
3 1 2
|
||
|
||
|
||
Output
|
||
|
||
YES
|
||
|
||
|
||
Input
|
||
|
||
2
|
||
galileo galilei
|
||
nicolaus copernicus
|
||
2 1
|
||
|
||
|
||
Output
|
||
|
||
YES
|
||
|
||
|
||
Input
|
||
|
||
10
|
||
rean schwarzer
|
||
fei claussell
|
||
alisa reinford
|
||
eliot craig
|
||
laura arseid
|
||
jusis albarea
|
||
machias regnitz
|
||
sara valestin
|
||
emma millstein
|
||
gaius worzel
|
||
1 2 3 4 5 6 7 8 9 10
|
||
|
||
|
||
Output
|
||
|
||
NO
|
||
|
||
|
||
Input
|
||
|
||
10
|
||
rean schwarzer
|
||
fei claussell
|
||
alisa reinford
|
||
eliot craig
|
||
laura arseid
|
||
jusis albarea
|
||
machias regnitz
|
||
sara valestin
|
||
emma millstein
|
||
gaius worzel
|
||
2 4 9 6 5 7 1 3 8 10
|
||
|
||
|
||
Output
|
||
|
||
YES
|
||
|
||
Note
|
||
|
||
In example 1 and 2, we have 3 people: tourist, Petr and me (cgy4ever). You can see that whatever handle is chosen, I must be the first, then tourist and Petr must be the last.
|
||
|
||
In example 3, if Copernicus uses \"copernicus\" as his handle, everything will be alright.
|
||
|
||
Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
|
||
```python
|
||
def MIN(t):
|
||
if t[0] < t[1]:
|
||
return t[0]
|
||
else:
|
||
return t[1]
|
||
def MAX(t):
|
||
if t[0] > t[1]:
|
||
return t[0]
|
||
else:
|
||
return t[1]
|
||
n = input()
|
||
|
||
name = []
|
||
while n:
|
||
n -= 1
|
||
a,b = raw_input().split()
|
||
name.append([a,b])
|
||
p = map(int,raw_input().split())
|
||
|
||
Flag = True
|
||
Tmp = MIN(name[p[0]-1])
|
||
for i in xrange(1,len(p)):
|
||
if Tmp >= MIN(name[p[i]-1]):
|
||
if Tmp >= MAX(name[p[i]-1]):
|
||
Flag = False
|
||
break
|
||
else:
|
||
Tmp = MAX(name[p[i]-1])
|
||
else:
|
||
Tmp = MIN(name[p[i]-1])
|
||
|
||
if Flag == True:
|
||
print \"YES\"
|
||
else:
|
||
print \"NO\"
|
||
```
|
||
|
||
|
||
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
|
||
import random
|
||
import re
|
||
from bootcamp import Basebootcamp
|
||
|
||
def solve_handle_order(names, p_list):
|
||
if not names or len(p_list) != len(names):
|
||
return "NO"
|
||
n = len(names)
|
||
p = p_list
|
||
Flag = True
|
||
current_user = names[p[0]-1]
|
||
a, b = current_user
|
||
Tmp = a if a < b else b
|
||
for i in range(1, len(p)):
|
||
current_user = names[p[i]-1]
|
||
a, b = current_user
|
||
current_min = a if a < b else b
|
||
current_max = b if a < b else a
|
||
if Tmp >= current_min:
|
||
if Tmp >= current_max:
|
||
Flag = False
|
||
break
|
||
else:
|
||
Tmp = current_max
|
||
else:
|
||
Tmp = current_min
|
||
return "YES" if Flag else "NO"
|
||
|
||
def generate_random_string(min_length=1, max_length=50):
|
||
length = random.randint(min_length, max_length)
|
||
return ''.join(random.choice('abcdefghijklmnopqrstuvwxyz') for _ in range(length))
|
||
|
||
class Cdesigntutorialmakeitnondeterministicbootcamp(Basebootcamp):
|
||
def __init__(self, min_n=3, max_n=10, solvable=None, max_attempts=1000):
|
||
super().__init__() # 显式调用父类初始化
|
||
self.min_n = min_n
|
||
self.max_n = max_n
|
||
self.solvable = solvable
|
||
self.max_attempts = max_attempts
|
||
|
||
def case_generator(self):
|
||
for _ in range(self.max_attempts):
|
||
# 保证n至少为1
|
||
n = random.randint(max(1, self.min_n), self.max_n)
|
||
|
||
names = []
|
||
all_names = set()
|
||
valid = True
|
||
|
||
# 为每个用户生成唯一的名字对
|
||
for _ in range(n):
|
||
attempt = 0
|
||
while True:
|
||
first = generate_random_string(1, 50)
|
||
last = generate_random_string(1, 50)
|
||
if first != last and first not in all_names and last not in all_names:
|
||
all_names.update([first, last])
|
||
names.append((first, last))
|
||
break
|
||
attempt += 1
|
||
if attempt > 100:
|
||
valid = False
|
||
break
|
||
if not valid:
|
||
break
|
||
if not valid:
|
||
continue
|
||
|
||
# 生成有效排列p
|
||
p = list(range(1, n+1))
|
||
random.shuffle(p)
|
||
answer = solve_handle_order(names, p)
|
||
|
||
# 根据solvable参数筛选案例
|
||
if self.solvable is None:
|
||
return {'n': n, 'names': names, 'p': p}
|
||
elif (self.solvable and answer == "YES") or (not self.solvable and answer == "NO"):
|
||
return {'n': n, 'names': names, 'p': p}
|
||
|
||
# 备用方案：生成确定性的可解/不可解案例
|
||
if self.solvable:
|
||
return {'n': 3, 'names': [('a','z'), ('b','y'), ('c','x')], 'p': [3,2,1]}
|
||
else:
|
||
return {'n': 3, 'names': [('z','a'), ('y','b'), ('x','c')], 'p': [1,2,3]}
|
||
|
||
@staticmethod
|
||
def prompt_func(question_case):
|
||
input_lines = [str(question_case['n'])]
|
||
for first, last in question_case['names']:
|
||
input_lines.append(f"{first} {last}")
|
||
input_lines.append(' '.join(map(str, question_case['p'])))
|
||
example_input = '\n'.join(input_lines)
|
||
|
||
prompt = (
|
||
"给定n个人，每个人可以选择使用名字或姓氏作为handle。\n"
|
||
"判断是否存在一种选择方式，使得按handle的字典序结果恰好等于给定的排列p。\n\n"
|
||
"输入格式：\n"
|
||
f"- 首行：n（人数）\n"
|
||
f"- 接下来n行：每行两个小写字母组成的字符串\n"
|
||
f"- 最后一行：排列p（1-based索引）\n\n"
|
||
"示例：\n"
|
||
"输入：\n3\na b\nc d\ne f\n3 1 2\n输出：YES\n\n"
|
||
"当前问题输入：\n"
|
||
f"{example_input}\n\n"
|
||
"请将最终答案放在[answer]和[/answer]标签之间。"
|
||
)
|
||
return prompt
|
||
|
||
@staticmethod
|
||
def extract_output(output):
|
||
# 支持多格式匹配（包含换行和大小写）
|
||
matches = re.findall(r'\[answer\s*\]\s*(.*?)\s*\[\s*/answer\s*\]', output, re.IGNORECASE | re.DOTALL)
|
||
if not matches:
|
||
return None
|
||
last_match = matches[-1].strip().upper()
|
||
return last_match if last_match in {"YES", "NO"} else None
|
||
|
||
@classmethod
|
||
def _verify_correction(cls, solution, identity):
|
||
# 验证前先进行基本清洗
|
||
cleaned_solution = str(solution).strip().upper()
|
||
return cleaned_solution == solve_handle_order(identity['names'], identity['p'])
|
||
|
||
# 测试用例示例
|
||
if __name__ == "__main__":
|
||
bootcamp = Cdesigntutorialmakeitnondeterministicbootcamp(solvable=True)
|
||
case = bootcamp.case_generator()
|
||
print("生成案例:", case)
|
||
print("问题描述:", bootcamp.prompt_func(case)[:200] + "...")
|
||
print("验证结果:", bootcamp._verify_correction("YES", case))
|