OI-Problems/ABC406F.cpp at main · WRT-Partisan/OI-Problems · GitHub

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/*
 * @FilePath: \c++\ABC406F.cpp
 * @Author: WRT_Partisan
 * @Date: 2025-10-04 17:37:00
 */
#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef unsigned long long ull;
#define mid ((l + r) >> 1)
struct Edge
{
    int u, v;
} e[300005];
int n, u, v, q, op, w[300005], sum, d[300005], s[300005], cnt, id[300005];
vector<int> mp[300005];
struct Node
{
    int s[1200005];
    void pushup(int x) { s[x] = s[x << 1] + s[x << 1 | 1]; }
    void build(int x, int l, int r)
    {
        if (l == r)
            return s[x] = 1, void();
        build(x << 1, l, mid);
        build(x << 1 | 1, mid + 1, r);
        pushup(x);
    }
    int ask(int x, int l, int r, int ql, int qr)
    {
        if (ql <= l && r <= qr)
            return s[x];
        int res = 0;
        if (ql <= mid)
            res += ask(x << 1, l, mid, ql, qr);
        if (mid < qr)
            res += ask(x << 1 | 1, mid + 1, r, ql, qr);
        return res;
    }
    void update(int x, int l, int r, int p, int u)
    {
        if (l == r)
            return s[x] += u, void();
        if (p <= mid)
            update(x << 1, l, mid, p, u);
        else
            update(x << 1 | 1, mid + 1, r, p, u);
        pushup(x);
    }
} tree;
void dfs(int x, int f)
{
    id[x] = ++cnt, d[x] = d[f] + 1, s[x] = 1;
    for (auto i : mp[x])
        if (i != f)
            dfs(i, x), s[x] += s[i];
}
signed main()
{
    cin.tie(0)->sync_with_stdio(0);
    cin >> n, sum = n;
    fill(w + 1, w + n + 1, 1);
    for (int i = 1; i < n; i++)
        cin >> e[i].u >> e[i].v, mp[e[i].u].push_back(e[i].v), mp[e[i].v].push_back(e[i].u);
    dfs(1, 0);
    tree.build(1, 1, cnt);
    for (int i = 1; i < n; i++)
        if (d[e[i].u] < d[e[i].v])
            swap(e[i].u, e[i].v);
    cin >> q;
    while (q--)
    {
        cin >> op >> u;
        if (op == 1)
            cin >> v, tree.update(1, 1, cnt, id[u], v), sum += v;
        else
            cout << abs(tree.ask(1, 1, cnt, id[e[u].u], id[e[u].u] + s[e[u].u] - 1) * 2 - sum) << "\n";
    }
    return 0;
}