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292 lines
9.4 KiB
Python
Executable file
292 lines
9.4 KiB
Python
Executable file
"""#
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### 谜题描述
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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.
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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.
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This tree supports two types of queries:
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* \"1 x val\" — val is added to the value of node x;
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* \"2 x\" — print the current value of node x.
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In order to help Iahub understand the tree better, you must answer m queries of the preceding type.
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Input
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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.
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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.
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Output
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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.
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Examples
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Input
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5 5
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1 2 1 1 2
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1 2
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1 3
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2 4
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2 5
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1 2 3
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1 1 2
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2 1
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2 2
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2 4
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Output
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3
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3
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0
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Note
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The values of the nodes are [1, 2, 1, 1, 2] at the beginning.
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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].
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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].
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You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
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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 int c = 2e5 + 5;
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int strength[c], level[c], size[c], visited[c], tree_even[4 * c],
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tree_odd[4 * c], M;
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std::vector<int> v[c], preorder;
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map<int, int> Map;
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int dfs(int vertex, int lev) {
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if (visited[vertex]) return 0;
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visited[vertex] = 1;
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level[vertex] = lev;
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size[vertex] = 1;
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preorder.push_back(vertex);
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Map[vertex] = (int)preorder.size() - 1;
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for (int i = 0; i < (int)v[vertex].size(); i++) {
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size[vertex] += dfs(v[vertex][i], lev + 1);
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}
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return size[vertex];
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}
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void update(int index, int s, int e, int l, int r, int val) {
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if (e < l || s > r) return;
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if (s >= l && e <= r) {
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if (M == 0)
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tree_even[index] += val;
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else
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tree_odd[index] += val;
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return;
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}
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int mid = (s + e) / 2;
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update(2 * index, s, mid, l, r, val);
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update(2 * index + 1, mid + 1, e, l, r, val);
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return;
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}
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int sum;
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int query(int index, int l, int r, int i) {
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if (i < l || i > r) return 0;
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if (l <= i && i <= r) {
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if (M == 0)
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sum += tree_even[index] - tree_odd[index];
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else
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sum += tree_odd[index] - tree_even[index];
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}
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if (l == r) return 0;
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int mid = (l + r) / 2;
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query(2 * index, l, mid, i);
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query(2 * index + 1, mid + 1, r, i);
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return 0;
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}
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int main() {
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ios_base::sync_with_stdio(false);
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cin.tie(NULL), cout.tie(NULL);
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int n, m, v1, v2, type, node, upd, start;
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cin >> n >> m;
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for (int i = 1; i < n + 1; i++) cin >> strength[i];
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for (int i = 0; i < n - 1; i++) {
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cin >> v1 >> v2;
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v[v1].push_back(v2);
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v[v2].push_back(v1);
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}
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preorder.push_back(0);
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dfs(1, 0);
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while (m--) {
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cin >> type;
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if (type == 1) {
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cin >> node >> upd;
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start = Map[node];
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M = level[node] % 2;
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update(1, 1, n, start, start + size[node] - 1, upd);
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} else {
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cin >> node;
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start = Map[node];
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M = level[node] % 2;
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sum = strength[node];
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query(1, 1, n, start);
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cout << sum << \"\n\";
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}
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}
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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 re
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import random
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from collections import deque
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class Epropagatingtreebootcamp(Basebootcamp):
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def __init__(self, n=5, m=5):
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self.n = n
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self.m = m
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def case_generator(self):
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n = self.n
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edges, parent = self.generate_tree(n)
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level = self.compute_levels(parent, n)
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children = self.build_children_dict(parent)
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subtree_nodes = {x: self.get_subtree_nodes(x, children) for x in range(1, n+1)}
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initial_a = [random.randint(1, 1000) for _ in range(n)]
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current_values = initial_a.copy()
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values_history = [current_values.copy()]
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queries = []
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correct_outputs = []
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# Generate all queries first
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for _ in range(self.m):
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if len(correct_outputs) < self.m // 2 or random.random() < 0.5:
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# Generate type 1 query
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x = random.randint(1, n)
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val = random.randint(1, 1000)
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queries.append(('1', x, val))
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# Record state before applying
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values_history.append(current_values.copy())
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# Apply changes
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x_level = level[x]
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for y in subtree_nodes[x]:
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delta = val * ((-1) ** (level[y] - x_level))
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current_values[y-1] += delta
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else:
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# Generate type 2 query
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x = random.randint(1, n)
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queries.append(('2', x))
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correct_outputs.append(current_values[x-1])
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values_history.append(current_values.copy())
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# Ensure at least one type 2 query and correct outputs
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if not correct_outputs:
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# Find last type 1 query to replace
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for i in reversed(range(len(queries))):
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if queries[i][0] == '1':
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# Get state before this query
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prev_values = values_history[i]
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x = random.randint(1, n)
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# Replace with type 2 query
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queries[i] = ('2', x)
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correct_outputs.insert(i - sum(1 for q in queries[:i] if q[0] == '2'), prev_values[x-1])
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break
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# Format queries to strings
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formatted_queries = []
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for q in queries:
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if q[0] == '1':
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formatted_queries.append(f'1 {q[1]} {q[2]}')
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else:
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formatted_queries.append(f'2 {q[1]}')
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case = {
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'n': n,
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'm': self.m,
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'a': initial_a,
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'edges': [[u, v] for u, v in edges],
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'queries': formatted_queries,
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'correct_outputs': correct_outputs
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}
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return case
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@staticmethod
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def generate_tree(n):
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if n == 1:
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return [], {1: None}
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parent = {1: None}
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edges = []
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available = [1]
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for i in range(2, n+1):
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p = random.choice(available)
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parent[i] = p
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edges.append((p, i))
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available.append(i)
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return edges, parent
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@staticmethod
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def compute_levels(parent, n):
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level = {1: 0}
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for i in range(2, n+1):
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level[i] = level[parent[i]] + 1
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return level
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@staticmethod
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def build_children_dict(parent):
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children = {}
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for child in parent:
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p = parent.get(child)
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if p is not None:
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children.setdefault(p, []).append(child)
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return children
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@staticmethod
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def get_subtree_nodes(x, children):
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subtree = []
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stack = [x]
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while stack:
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node = stack.pop()
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subtree.append(node)
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stack.extend(reversed(children.get(node, []))) # Maintain order
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return subtree
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@staticmethod
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def prompt_func(question_case) -> str:
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input_lines = [
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f"{question_case['n']} {question_case['m']}",
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' '.join(map(str, question_case['a']))
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]
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input_lines += [' '.join(map(str, e)) for e in question_case['edges']]
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input_lines += question_case['queries']
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joined_input = '\n'.join(input_lines)
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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.
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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.
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Input:
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{joined_input}
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Provide your answers for type 2 queries in order, each enclosed in [answer] tags:"""
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return problem_text
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@staticmethod
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def extract_output(output):
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answers = []
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matches = re.findall(r'\[answer\]\s*(-?\d+)\s*\[/answer\]', output, re.IGNORECASE)
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return list(map(int, matches)) if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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expected = identity.get('correct_outputs', [])
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return isinstance(solution, list) and solution == expected
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