Submission #593766

#	Time	Username	Problem	Language	Result	Execution time	Memory
593766		bogdanvladmihai	Ball Machine (BOI13_ballmachine)	C++14	16.11 / 100	148 ms	33292 KiB

ballmachine

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

/***
Given a rooted tree, you have to support 2 types of operations:
* paint the root of a full black subtree in black
* check where the path from a black node to the root becomes white

I can do the erases using binary lifting:
* if the 2^k ancestor of node u is black, than all the nodes between them are black
* if not, you can go down the tree

Add operation:
* I will always want to add in the same exact order
* So I will have a vector of 1 and 0, and I want to find the LAST 0

***/

const int LG = 17;
const int MAXN = 1000 * 100;

vector<int> g[MAXN + 1], order[MAXN + 1];
int dad[MAXN + 1], minValue[MAXN + 1], who[MAXN + 1], tin[MAXN + 1];
int ancestor[LG + 1][MAXN + 1];
bool black[MAXN + 1];
priority_queue<int> fr;

void dfs(int u) {
    vector<pair<int, int>> children;

    for (const int &v : g[u]) {
        dfs(v);

        minValue[u] = min(minValue[u], minValue[v]);
        children.emplace_back(minValue[v], v);
    }

    sort(children.begin(), children.end());
    for (const auto &p : children) {
        order[u].push_back(p.second);
    }
}

void dfsOrder(int u) {
    static int t = 0;

    tin[u] = t++;
    who[t - 1] = u;

    for (const int &v : order[u]) {
        dfsOrder(v);
    }
}

int add() {
    int u = who[fr.top()];
    fr.pop();

    black[u] = true;
    return u;
}

int erase(int u) {
    int ans = 0;
    for (int l = LG; l >= 0; l--) {
        if (black[ancestor[l][u]]) {
            u = ancestor[l][u];
            ans += (1 << l);
        }
    }

    fr.push(tin[u]);
    black[u] = false;
    return ans;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    int n, q; cin >> n >> q;

    int root = -1;
    for (int i = 1; i <= n; i++) {
        cin >> dad[i];

        if (dad[i] != 0) {
            g[dad[i]].push_back(i);
        } else {
            root = i;
        }
    }

    for (int i = 1; i <= n; i++) {
        minValue[i] = i;
        fr.push(i - 1);
    }

    dfs(root);
    dfsOrder(root);

    for (int u = 1; u <= n; u++) {
        ancestor[0][u] = dad[u];
    }

    for (int k = 1; k <= LG; k++) {
        for (int u = 1; u <= n; u++) {
            ancestor[k][u] = ancestor[k - 1][ancestor[k - 1][u]];
        }
    }

    for (int i = 0; i < q; i++) {
        int op, k; cin >> op >> k;

        if (op == 1) {
            int u = -1;
            for (int j = 0; j < k; j++) {
                u = add();
            }

            cout << u << "\n";
        } else {
            cout << erase(k) << "\n";
        }
    }

    return 0;
}

• Subtask 1: 0 / 25
• Subtask 2: 6.428571428571429 / 30
• Subtask 3: 8.571428571428571 / 40
• Subtask 4: 1.1111111111111112 / 5