init-commit

internbootcamp/bootcamp/epropagatingtree/epropagatingtree.py

@@ -0,0 +1,292 @@
+"""# 
+
+### 谜题描述
+Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.
+
+This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.
+
+This tree supports two types of queries:
+
+  * \"1 x val\" — val is added to the value of node x; 
+  * \"2 x\" — print the current value of node x. 
+
+
+
+In order to help Iahub understand the tree better, you must answer m queries of the preceding type.
+
+Input
+
+The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers vi and ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.
+
+Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.
+
+Output
+
+For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.
+
+Examples
+
+Input
+
+5 5
+1 2 1 1 2
+1 2
+1 3
+2 4
+2 5
+1 2 3
+1 1 2
+2 1
+2 2
+2 4
+
+
+Output
+
+3
+3
+0
+
+Note
+
+The values of the nodes are [1, 2, 1, 1, 2] at the beginning.
+
+Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1, - 2, - 1].
+
+Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3, - 1, 0, 1].
+
+You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
+
+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;
+const int c = 2e5 + 5;
+int strength[c], level[c], size[c], visited[c], tree_even[4 * c],
+    tree_odd[4 * c], M;
+std::vector<int> v[c], preorder;
+map<int, int> Map;
+int dfs(int vertex, int lev) {
+  if (visited[vertex]) return 0;
+  visited[vertex] = 1;
+  level[vertex] = lev;
+  size[vertex] = 1;
+  preorder.push_back(vertex);
+  Map[vertex] = (int)preorder.size() - 1;
+  for (int i = 0; i < (int)v[vertex].size(); i++) {
+    size[vertex] += dfs(v[vertex][i], lev + 1);
+  }
+  return size[vertex];
+}
+void update(int index, int s, int e, int l, int r, int val) {
+  if (e < l || s > r) return;
+  if (s >= l && e <= r) {
+    if (M == 0)
+      tree_even[index] += val;
+    else
+      tree_odd[index] += val;
+    return;
+  }
+  int mid = (s + e) / 2;
+  update(2 * index, s, mid, l, r, val);
+  update(2 * index + 1, mid + 1, e, l, r, val);
+  return;
+}
+int sum;
+int query(int index, int l, int r, int i) {
+  if (i < l || i > r) return 0;
+  if (l <= i && i <= r) {
+    if (M == 0)
+      sum += tree_even[index] - tree_odd[index];
+    else
+      sum += tree_odd[index] - tree_even[index];
+  }
+  if (l == r) return 0;
+  int mid = (l + r) / 2;
+  query(2 * index, l, mid, i);
+  query(2 * index + 1, mid + 1, r, i);
+  return 0;
+}
+int main() {
+  ios_base::sync_with_stdio(false);
+  cin.tie(NULL), cout.tie(NULL);
+  int n, m, v1, v2, type, node, upd, start;
+  cin >> n >> m;
+  for (int i = 1; i < n + 1; i++) cin >> strength[i];
+  for (int i = 0; i < n - 1; i++) {
+    cin >> v1 >> v2;
+    v[v1].push_back(v2);
+    v[v2].push_back(v1);
+  }
+  preorder.push_back(0);
+  dfs(1, 0);
+  while (m--) {
+    cin >> type;
+    if (type == 1) {
+      cin >> node >> upd;
+      start = Map[node];
+      M = level[node] % 2;
+      update(1, 1, n, start, start + size[node] - 1, upd);
+    } else {
+      cin >> node;
+      start = Map[node];
+      M = level[node] % 2;
+      sum = strength[node];
+      query(1, 1, n, start);
+      cout << sum << \"\n\";
+    }
+  }
+  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 re
+import random
+from collections import deque
+
+class Epropagatingtreebootcamp(Basebootcamp):
+    def __init__(self, n=5, m=5):
+        self.n = n
+        self.m = m
+
+    def case_generator(self):
+        n = self.n
+        edges, parent = self.generate_tree(n)
+        level = self.compute_levels(parent, n)
+        children = self.build_children_dict(parent)
+        subtree_nodes = {x: self.get_subtree_nodes(x, children) for x in range(1, n+1)}
+        initial_a = [random.randint(1, 1000) for _ in range(n)]
+        
+        current_values = initial_a.copy()
+        values_history = [current_values.copy()]
+        queries = []
+        correct_outputs = []
+        
+        # Generate all queries first
+        for _ in range(self.m):
+            if len(correct_outputs) < self.m // 2 or random.random() < 0.5:
+                # Generate type 1 query
+                x = random.randint(1, n)
+                val = random.randint(1, 1000)
+                queries.append(('1', x, val))
+                
+                # Record state before applying
+                values_history.append(current_values.copy())
+                # Apply changes
+                x_level = level[x]
+                for y in subtree_nodes[x]:
+                    delta = val * ((-1) ** (level[y] - x_level))
+                    current_values[y-1] += delta
+            else:
+                # Generate type 2 query
+                x = random.randint(1, n)
+                queries.append(('2', x))
+                correct_outputs.append(current_values[x-1])
+                values_history.append(current_values.copy())
+        
+        # Ensure at least one type 2 query and correct outputs
+        if not correct_outputs:
+            # Find last type 1 query to replace
+            for i in reversed(range(len(queries))):
+                if queries[i][0] == '1':
+                    # Get state before this query
+                    prev_values = values_history[i]
+                    x = random.randint(1, n)
+                    # Replace with type 2 query
+                    queries[i] = ('2', x)
+                    correct_outputs.insert(i - sum(1 for q in queries[:i] if q[0] == '2'), prev_values[x-1])
+                    break
+        
+        # Format queries to strings
+        formatted_queries = []
+        for q in queries:
+            if q[0] == '1':
+                formatted_queries.append(f'1 {q[1]} {q[2]}')
+            else:
+                formatted_queries.append(f'2 {q[1]}')
+        
+        case = {
+            'n': n,
+            'm': self.m,
+            'a': initial_a,
+            'edges': [[u, v] for u, v in edges],
+            'queries': formatted_queries,
+            'correct_outputs': correct_outputs
+        }
+        return case
+
+    @staticmethod
+    def generate_tree(n):
+        if n == 1:
+            return [], {1: None}
+        parent = {1: None}
+        edges = []
+        available = [1]
+        for i in range(2, n+1):
+            p = random.choice(available)
+            parent[i] = p
+            edges.append((p, i))
+            available.append(i)
+        return edges, parent
+
+    @staticmethod
+    def compute_levels(parent, n):
+        level = {1: 0}
+        for i in range(2, n+1):
+            level[i] = level[parent[i]] + 1
+        return level
+
+    @staticmethod
+    def build_children_dict(parent):
+        children = {}
+        for child in parent:
+            p = parent.get(child)
+            if p is not None:
+                children.setdefault(p, []).append(child)
+        return children
+
+    @staticmethod
+    def get_subtree_nodes(x, children):
+        subtree = []
+        stack = [x]
+        while stack:
+            node = stack.pop()
+            subtree.append(node)
+            stack.extend(reversed(children.get(node, [])))  # Maintain order
+        return subtree
+
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        input_lines = [
+            f"{question_case['n']} {question_case['m']}",
+            ' '.join(map(str, question_case['a']))
+        ]
+        input_lines += [' '.join(map(str, e)) for e in question_case['edges']]
+        input_lines += question_case['queries']
+        joined_input = '\n'.join(input_lines)
+        
+        problem_text = f"""Iahub discovered a propagating tree with a special property. When a value is added to a node, it propagates to children with alternating signs. 
+
+You must process the following queries and output answers for type 2 queries. Format each answer on a separate line within [answer] and [/answer] tags.
+
+Input:
+{joined_input}
+
+Provide your answers for type 2 queries in order, each enclosed in [answer] tags:"""
+        return problem_text
+
+    @staticmethod
+    def extract_output(output):
+        answers = []
+        matches = re.findall(r'\[answer\]\s*(-?\d+)\s*\[/answer\]', output, re.IGNORECASE)
+        return list(map(int, matches)) if matches else None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        expected = identity.get('correct_outputs', [])
+        return isinstance(solution, list) and solution == expected