Editorial for Goblins

Use this editorial only when stuck, and do not copy-paste code from it.

Please be respectful to the problem author and editorialist. Submitting an official solution before solving the problem yourself is a bannable offence.

Approach

The superiors form a tree rooted at goblin $0$ . For a query at goblin $x$ , start at $x$ 's parent and walk toward the root, subtracting each goblin's power from Indra's budget $p$ , counting how many superiors can be defeated before the next one would exceed the remaining budget.

A naive walk is too slow for $n, q \le 10^5$ . Precompute binary lifting tables:

$up[v][j]$ : the $2^j$ -th superior of $v$ (or nonexistent)
$jump\_sum[v][j]$ : the sum of powers of the $2^j$ goblins starting at $v$ and walking up

Then each query greedily takes the largest jumps whose power sum fits in the remaining budget. This answers every query in $O(\log n)$ after $O(n \log n)$ preprocessing.

Solution (C++)

Code 1

#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, q;
    long long indra;
    cin >> n >> q >> indra;

    vector<long long> power(n);
    for (int i = 0; i < n; i++) {
        cin >> power[i];
    }

    vector<vector<int> > adj(n);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    vector<int> parent(n, -1);
    stack<int> st;
    st.push(0);
    parent[0] = -2;
    while (!st.empty()) {
        int v = st.top();
        st.pop();
        for (size_t i = 0; i < adj[v].size(); i++) {
            int to = adj[v][i];
            if (parent[to] != -1) {
                continue;
            }
            parent[to] = v;
            st.push(to);
        }
    }
    parent[0] = -1;

    const int LOGN = 18;
    const long long INF = (1LL << 62);
    vector<vector<int> > up(n, vector<int>(LOGN, -1));
    vector<vector<long long> > jump_sum(n, vector<long long>(LOGN, INF));

    for (int i = 0; i < n; i++) {
        up[i][0] = parent[i];
        jump_sum[i][0] = power[i];
    }
    for (int j = 1; j < LOGN; j++) {
        for (int i = 0; i < n; i++) {
            int mid = up[i][j - 1];
            if (mid == -1) {
                up[i][j] = -1;
                jump_sum[i][j] = INF;
            } else {
                up[i][j] = up[mid][j - 1];
                if (jump_sum[i][j - 1] > INF / 2 || jump_sum[mid][j - 1] > INF / 2) {
                    jump_sum[i][j] = INF;
                } else {
                    jump_sum[i][j] = jump_sum[i][j - 1] + jump_sum[mid][j - 1];
                }
            }
        }
    }

    for (int qi = 0; qi < q; qi++) {
        int at;
        cin >> at;
        int v = parent[at];
        long long budget = indra;
        int ans = 0;
        for (int j = LOGN - 1; j >= 0; j--) {
            if (v == -1) {
                break;
            }
            if (jump_sum[v][j] <= budget) {
                budget -= jump_sum[v][j];
                ans += 1 << j;
                v = up[v][j];
            }
        }
        cout << ans << "\n";
    }
}

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Editorial for Goblins

Approach

Solution (C++)

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