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internbootcamp/bootcamp/fboboniuandstring/fboboniuandstring.py

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+"""# 
+
+### 谜题描述
+Boboniu defines BN-string as a string s of characters 'B' and 'N'.
+
+You can perform the following operations on the BN-string s:
+
+  * Remove a character of s. 
+  * Remove a substring \"BN\" or \"NB\" of s. 
+  * Add a character 'B' or 'N' to the end of s. 
+  * Add a string \"BN\" or \"NB\" to the end of s. 
+
+
+
+Note that a string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
+
+Boboniu thinks that BN-strings s and t are similar if and only if:
+
+  * |s|=|t|. 
+  * There exists a permutation p_1, p_2, …, p_{|s|} such that for all i (1≤ i≤ |s|), s_{p_i}=t_i. 
+
+
+
+Boboniu also defines dist(s,t), the distance between s and t, as the minimum number of operations that makes s similar to t.
+
+Now Boboniu gives you n non-empty BN-strings s_1,s_2,…, s_n and asks you to find a non-empty BN-string t such that the maximum distance to string s is minimized, i.e. you need to minimize max_{i=1}^n dist(s_i,t).
+
+Input
+
+The first line contains a single integer n (1≤ n≤ 3⋅ 10^5).
+
+Each of the next n lines contains a string s_i (1≤ |s_i| ≤ 5⋅ 10^5). It is guaranteed that s_i only contains 'B' and 'N'. The sum of |s_i| does not exceed 5⋅ 10^5.
+
+Output
+
+In the first line, print the minimum max_{i=1}^n dist(s_i,t).
+
+In the second line, print the suitable t.
+
+If there are several possible t's, you can print any.
+
+Examples
+
+Input
+
+
+3
+B
+N
+BN
+
+
+Output
+
+
+1
+BN
+
+
+Input
+
+
+10
+N
+BBBBBB
+BNNNBBNBB
+NNNNBNBNNBNNNBBN
+NBNBN
+NNNNNN
+BNBNBNBBBBNNNNBBBBNNBBNBNBBNBBBBBBBB
+NNNNBN
+NBBBBBBBB
+NNNNNN
+
+
+Output
+
+
+12
+BBBBBBBBBBBBNNNNNNNNNNNN
+
+
+Input
+
+
+8
+NNN
+NNN
+BBNNBBBN
+NNNBNN
+B
+NNN
+NNNNBNN
+NNNNNNNNNNNNNNNBNNNNNNNBNB
+
+
+Output
+
+
+12
+BBBBNNNNNNNNNNNN
+
+
+Input
+
+
+3
+BNNNBNNNNBNBBNNNBBNNNNBBBBNNBBBBBBNBBBBBNBBBNNBBBNBNBBBN
+BBBNBBBBNNNNNBBNBBBNNNBB
+BBBBBBBBBBBBBBNBBBBBNBBBBBNBBBBNB
+
+
+Output
+
+
+12
+BBBBBBBBBBBBBBBBBBBBBBBBBBNNNNNNNNNNNN
+
+Note
+
+In the first example dist(B,BN)=dist(N,BN)=1, dist(BN,BN)=0. So the maximum distance is 1.
+
+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;
+#pragma GCC target(\"avx2\")
+#pragma GCC optimization(\"O3\")
+#pragma GCC optimization(\"unroll-loops\")
+const long long INFll = 1e18;
+const int INFint = 1e9;
+const long long MOD = 1e9 + 9;
+void needForSpeed() {
+  ios_base::sync_with_stdio(0);
+  cin.tie(0);
+  cout.tie(0);
+}
+bool isPrime(long long a) {
+  if (a < 2) return 0;
+  for (long long i = 2; i <= sqrt(a); i++)
+    if (a % i == 0) return 0;
+  return 1;
+}
+long long binpow(long long base, long long pow, long long mod = INFll) {
+  if (pow == 0)
+    return 1;
+  else if (pow % 2)
+    return base * binpow(base, pow - 1, mod) % mod;
+  else {
+    long long p = binpow(base, pow / 2, mod);
+    return p * p % mod;
+  }
+}
+template <class T>
+void PR_VEC(const vector<T> &vec) {
+  cout << \"size(\" << vec.size() << \"):[ '\";
+  cout << vec[0] << \"'\";
+  for (int i = 1; i < vec.size(); ++i) cout << \" , '\" << vec[i] << \"'\";
+  cout << \" ]\n\";
+}
+int solve() {
+  int n;
+  cin >> n;
+  vector<pair<int, int>> dots(n);
+  for (int i = 0; i < n; i++) {
+    string s;
+    cin >> s;
+    int x = 0, y = 0;
+    for (int j = 0; j < s.length(); j++) {
+      if (s[j] == 'B')
+        ++x;
+      else
+        ++y;
+    }
+    dots[i] = {x, y};
+  }
+  int l = -1, r = 10000000;
+  pair<int, int> final_dot = {0, 0};
+  while (r - l > 1) {
+    int mid = (l + r) / 2;
+    int minx = 0, miny = 0, maxx = 10000000, maxy = 10000000;
+    for (int i = 0; i < n; i++) {
+      minx = max(minx, dots[i].first - mid);
+      maxx = min(maxx, dots[i].first + mid);
+      miny = max(miny, dots[i].second - mid);
+      maxy = min(maxy, dots[i].second + mid);
+    }
+    int minXY = -10000000, maxXY = 10000000;
+    for (int i = 0; i < n; i++) {
+      minXY = max(minXY, dots[i].first - dots[i].second - mid);
+      maxXY = min(maxXY, dots[i].first - dots[i].second + mid);
+    }
+    bool may_be = (minx <= maxx && miny <= maxy && minXY <= maxXY);
+    if (minx - maxy > maxXY || maxx - miny < minXY) may_be = false;
+    if (may_be) {
+      final_dot = {minx, maxy};
+      if (final_dot.first - final_dot.second < minXY) {
+        int move = min(maxx - final_dot.first,
+                       minXY - (final_dot.first - final_dot.second));
+        final_dot.first += move;
+      }
+      if (final_dot.first - final_dot.second < minXY) {
+        int move = min(final_dot.second - miny,
+                       minXY - (final_dot.first - final_dot.second));
+        final_dot.second -= move;
+      }
+      r = mid;
+    } else
+      l = mid;
+  }
+  cout << r << \"\n\";
+  for (int i = 0; i < final_dot.first; i++) cout << 'B';
+  for (int i = 0; i < final_dot.second; i++) cout << 'N';
+  cout << \"\n\";
+  return 0;
+}
+int main() {
+  needForSpeed();
+  int tests = 1;
+  while (tests--) {
+    solve();
+  }
+  return 0;
+}
+```
+
+
+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
+from bootcamp import Basebootcamp
+
+def solve_bn_case(dots):
+    n = len(dots)
+    l = -1
+    r = 10**7
+    final_dot = (0, 0)
+    while r - l > 1:
+        mid = (l + r) // 2
+        minx = -10**7
+        maxx = 10**7
+        miny = -10**7
+        maxy = 10**7
+        minXY = -10**7
+        maxXY = 10**7
+        
+        for x, y in dots:
+            minx = max(minx, x - mid)
+            maxx = min(maxx, x + mid)
+            miny = max(miny, y - mid)
+            maxy = min(maxy, y + mid)
+            minXY = max(minXY, (x - y) - mid)
+            maxXY = min(maxXY, (x - y) + mid)
+        
+        may_be = (minx <= maxx) and (miny <= maxy) and (minXY <= maxXY)
+        if may_be:
+            lower_bound = minx - maxy
+            upper_bound = maxx - miny
+            if lower_bound > maxXY or upper_bound < minXY:
+                may_be = False
+        
+        if may_be:
+            x_t = minx
+            y_t = maxy
+            if (x_t - y_t) < minXY:
+                move = min(maxx - x_t, minXY - (x_t - y_t))
+                x_t += move
+                if (x_t - y_t) < minXY:
+                    move = min(y_t - miny, minXY - (x_t - y_t))
+                    y_t -= move
+            x_t = max(x_t, 0)
+            y_t = max(y_t, 0)
+            if x_t == 0 and y_t == 0:
+                x_t = 1
+                y_t = 0
+            final_dot = (x_t, y_t)
+            r = mid
+        else:
+            l = mid
+    return r, final_dot
+
+class Fboboniuandstringbootcamp(Basebootcamp):
+    def __init__(self, n=3, max_b=5, max_n=5):
+        self.n = n
+        self.max_b = max_b
+        self.max_n = max_n
+    
+    def case_generator(self):
+        dots = []
+        for _ in range(self.n):
+            x = random.randint(0, self.max_b)
+            y = random.randint(0, self.max_n)
+            while x + y == 0:
+                x = random.randint(0, self.max_b)
+                y = random.randint(0, self.max_n)
+            dots.append((x, y))
+        correct_d_max, (t_x, t_y) = solve_bn_case(dots)
+        strings = []
+        for x, y in dots:
+            strings.append('B' * x + 'N' * y)
+        return {
+            'n': self.n,
+            'strings': strings,
+            'dots': dots,
+            'correct_d_max': correct_d_max
+        }
+    
+    @staticmethod
+    def prompt_func(question_case):
+        input_lines = [str(question_case['n'])] + question_case['strings']
+        input_part = '\n'.join(input_lines)
+        prompt = f"""You need to solve a BN-string optimization problem. Find a non-empty BN-string t such that the maximum distance to given strings is minimized. 
+
+**Problem Details:**
+- Distance 'dist(s, t)' is the minimum operations to make s similar to t, where similarity requires equal length and possible permutation of characters.
+- Operations include adding/removing characters or substrings "BN"/"NB".
+
+**Input:**
+{input_part}
+
+**Output Format:**
+First line: the minimal maximum distance.
+Second line: the optimal BN-string t.
+
+Put your answer within [answer] and [/answer] tags. Example:
+[answer]
+3
+BNNBB
+[/answer]"""
+        return prompt
+    
+    @staticmethod
+    def extract_output(output):
+        import re
+        answer_blocks = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not answer_blocks:
+            return None
+        last_answer = answer_blocks[-1].strip()
+        lines = [line.strip() for line in last_answer.split('\n') if line.strip()]
+        if len(lines) < 2:
+            return None
+        d_str, t = lines[0], lines[1]
+        if not t or any(c not in {'B', 'N'} for c in t):
+            return None
+        try:
+            d = int(d_str)
+        except:
+            return None
+        return {'d': d, 't': t}
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        if not solution:
+            return False
+        try:
+            user_d = solution['d']
+            user_t = solution['t']
+        except:
+            return False
+        if not user_t or any(c not in {'B', 'N'} for c in user_t):
+            return False
+        x_t = user_t.count('B')
+        y_t = user_t.count('N')
+        if x_t + y_t == 0:
+            return False
+        correct_d = identity['correct_d_max']
+        max_dist = 0
+        for x_i, y_i in identity['dots']:
+            dx = abs(x_t - x_i)
+            dy = abs(y_t - y_i)
+            d_xy = abs((x_t - y_t) - (x_i - y_i))
+            current = max(dx, dy, d_xy)
+            if current > max_dist:
+                max_dist = current
+        return max_dist == correct_d and user_d == correct_d