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270 lines
8.4 KiB
Python
Executable file
270 lines
8.4 KiB
Python
Executable file
"""#
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### 谜题描述
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You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.
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Initially all vertices contain number 0. Then come q queries, each query has one of the two types:
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* The format of the query: 1 v x k. In response to the query, you need to add to the number at vertex v number x; to the numbers at the descendants of vertex v at distance 1, add x - k; and so on, to the numbers written in the descendants of vertex v at distance i, you need to add x - (i·k). The distance between two vertices is the number of edges in the shortest path between these vertices.
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* The format of the query: 2 v. In reply to the query you should print the number written in vertex v modulo 1000000007 (109 + 7).
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Process the queries given in the input.
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Input
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The first line contains integer n (1 ≤ n ≤ 3·105) — the number of vertices in the tree. The second line contains n - 1 integers p2, p3, ... pn (1 ≤ pi < i), where pi is the number of the vertex that is the parent of vertex i in the tree.
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The third line contains integer q (1 ≤ q ≤ 3·105) — the number of queries. Next q lines contain the queries, one per line. The first number in the line is type. It represents the type of the query. If type = 1, then next follow space-separated integers v, x, k (1 ≤ v ≤ n; 0 ≤ x < 109 + 7; 0 ≤ k < 109 + 7). If type = 2, then next follows integer v (1 ≤ v ≤ n) — the vertex where you need to find the value of the number.
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Output
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For each query of the second type print on a single line the number written in the vertex from the query. Print the number modulo 1000000007 (109 + 7).
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Examples
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Input
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3
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1 1
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3
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1 1 2 1
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2 1
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2 2
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Output
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2
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1
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Note
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You can read about a rooted tree here: 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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const int N = 300200;
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using namespace std;
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int n;
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int tim;
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int tin[N];
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int tout[N];
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int dep[N];
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int mod = 1e9 + 7;
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pair<int, int> t[4 * N];
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vector<int> v[N];
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void dfs(int x, int g) {
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tin[x] = tout[x] = ++tim;
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dep[tin[x]] = g;
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for (auto y : v[x]) {
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dfs(y, g - 1);
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tout[x] = tout[y];
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}
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}
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void push(int x) {
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t[x * 2].first += t[x].first;
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t[x * 2].second += t[x].second;
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t[x * 2 + 1].first += t[x].first;
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t[x * 2 + 1].second += t[x].second;
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t[x * 2].first %= mod;
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t[x * 2].second %= mod;
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t[x * 2 + 1].first %= mod;
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t[x * 2 + 1].second %= mod;
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t[x].first = t[x].second = 0;
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}
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void upd(int x, int l, int r, int tl, int tr, int f1, int f2) {
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if (tl > tr) return;
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if (l == tl && r == tr) {
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t[x].first = (t[x].first + f1) % mod;
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t[x].second = (t[x].second + f2) % mod;
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return;
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}
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push(x);
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int m = (l + r) / 2;
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upd(x * 2, l, m, tl, min(m, tr), f1, f2);
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upd(x * 2 + 1, m + 1, r, max(m + 1, tl), tr, f1, f2);
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}
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int get(int x, int l, int r, int g) {
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if (l == r) {
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int ans = (1ll * t[x].second * dep[l]) % mod;
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ans = (t[x].first + ans) % mod;
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return ans;
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}
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push(x);
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int m = (l + r) / 2;
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if (g <= m)
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return get(x * 2, l, m, g);
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else
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return get(x * 2 + 1, m + 1, r, g);
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}
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int main() {
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ios_base::sync_with_stdio(0);
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scanf(\"%d\", &n);
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for (int i = 2; i <= n; i++) {
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int x;
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scanf(\"%d\", &x);
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v[x].push_back(i);
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}
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dfs(1, n);
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int q;
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scanf(\"%d\", &q);
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for (int i = 1; i <= q; i++) {
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int t, v, x, k;
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scanf(\"%d\", &t);
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if (t == 1) {
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scanf(\"%d%d%d\", &v, &x, &k);
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long long f = 1ll * x - 1ll * dep[tin[v]] * k;
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f = f % mod + mod;
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upd(1, 1, n, tin[v], tout[v], f % mod, k);
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} else {
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scanf(\"%d\", &v);
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printf(\"%d\n\", get(1, 1, n, tin[v]));
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}
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}
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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 random
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import re
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from bootcamp import Basebootcamp
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mod = 10**9 + 7
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def build_adj(parents, n):
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adj = {i: [] for i in range(1, n+1)}
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for i in range(2, n+1):
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parent = parents[i-2]
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adj[parent].append(i)
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return adj
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def dfs(x, parent_adj, tin, tout, dep, current_time, current_g):
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current_time[0] += 1
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tin[x] = current_time[0]
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dep[tin[x]] = current_g
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for child in parent_adj.get(x, []):
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dfs(child, parent_adj, tin, tout, dep, current_time, current_g-1)
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tout[x] = current_time[0]
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def perform_dfs(n, parent_adj):
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tin = [0] * (n + 1)
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tout = [0] * (n + 1)
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dep = [0] * (n + 2) # tin values are 1-based
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current_time = [0]
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dfs(1, parent_adj, tin, tout, dep, current_time, n)
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return tin, tout, dep
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def process_queries_for_identity(queries, n, tin_dict, tout_dict, dep_list):
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a = [0] * (n + 2)
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b = [0] * (n + 2)
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expected_outputs = []
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for query in queries:
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if query['type'] == 1:
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v = query['v']
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x = query['x']
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k = query['k']
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tin_v = tin_dict[v]
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tout_v = tout_dict[v]
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f1 = (x - dep_list[tin_v] * k) % mod
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f2 = k % mod
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for u in range(1, n+1):
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u_tin = tin_dict[u]
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if tin_v <= u_tin <= tout_v:
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a[u_tin] = (a[u_tin] + f1) % mod
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b[u_tin] = (b[u_tin] + f2) % mod
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else:
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v = query['v']
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u_tin = tin_dict[v]
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res = (a[u_tin] + b[u_tin] * dep_list[u_tin]) % mod
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expected_outputs.append(res)
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return expected_outputs
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class Eonchangingtreebootcamp(Basebootcamp):
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def __init__(self, max_n=5, max_q=5):
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self.max_n = max_n
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self.max_q = max_q
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def case_generator(self):
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n = random.randint(1, self.max_n)
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parents = []
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if n > 1:
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parents = [random.randint(1, i-1) for i in range(2, n+1)]
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adj = build_adj(parents, n)
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tin, tout, dep = perform_dfs(n, adj)
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tin_dict = {x: tin[x] for x in range(1, n+1)}
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tout_dict = {x: tout[x] for x in range(1, n+1)}
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q = random.randint(1, self.max_q)
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queries = []
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for _ in range(q):
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if random.random() < 0.3 or not any(q.get('type') == 2 for q in queries):
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v = random.randint(1, n)
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queries.append({'type': 2, 'v': v})
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else:
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v = random.randint(1, n)
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x = random.randint(0, mod-1)
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k = random.randint(0, mod-1)
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queries.append({'type': 1, 'v': v, 'x': x, 'k': k})
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expected_outputs = process_queries_for_identity(queries, n, tin_dict, tout_dict, dep)
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new_queries = []
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output_idx = 0
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for q in queries:
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if q['type'] == 2:
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new_q = q.copy()
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new_q['expected'] = expected_outputs[output_idx]
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new_queries.append(new_q)
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output_idx += 1
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else:
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new_queries.append(q)
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return {
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'n': n,
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'parents': parents,
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'queries': new_queries,
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'tin_dict': tin_dict,
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'tout_dict': tout_dict,
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'dep_list': dep
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}
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@staticmethod
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def prompt_func(question_case):
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input_lines = [
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str(question_case['n']),
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' '.join(map(str, question_case['parents'])) if question_case['n'] > 1 else '',
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str(len(question_case['queries']))
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]
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for q in question_case['queries']:
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if q['type'] == 1:
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input_lines.append(f"1 {q['v']} {q['x']} {q['k']}")
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else:
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input_lines.append(f"2 {q['v']}")
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input_str = '\n'.join(input_lines)
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return f"""你正在解决一个树处理问题。处理所有查询并将每个类型2的答案放在[answer]和[/answer]之间。输入数据如下：
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{input_str}
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规则：
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1. 类型1查询为节点v及其后代按距离加值。
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2. 类型2查询输出节点值模10^9+7的结果。
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答案格式示例：
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[answer]结果1[/answer]
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[answer]结果2[/answer]"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output)
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try:
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return [int(m) % mod for m in matches] if matches else None
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except:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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expected = [q['expected'] for q in identity['queries'] if q['type'] == 2]
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return solution == expected if solution else False
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