Submission #1143134

#	Time	Username	Problem	Language	Result	Execution time	Memory
1143134		adimiclaus15	Triumphal arch (POI13_luk)	C++20	0 / 100	88 ms	18504 KiB

luk

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

static const int MAXN = 300000;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    // Edge case: if n == 1, only the capital exists, so no crews needed.
    if (n == 1) {
        cout << 0 << "\n";
        return 0;
    }

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

    // BFS from node 1 to find distances
    vector<int> dist(n+1, -1);
    dist[1] = 0;
    queue<int>q;
    q.push(1);

    while (!q.empty()) {
        int u = q.front(); 
        q.pop();
        for (int v : adj[u]) {
            if (dist[v] == -1) {
                dist[v] = dist[u] + 1;
                q.push(v);
            }
        }
    }

    // We won't build an arch for the capital (dist=0).
    // Collect frequency of distances for dist >= 1
    // Note: max distance in a tree can be up to (n-1) in a worst-case chain.
    // But we'll track only up to the maximum distance we actually see.
    int maxD = 0;
    for (int v = 2; v <= n; v++) {
        maxD = max(maxD, dist[v]);
    }

    vector<long long> freq(maxD+1, 0LL);
    for (int v = 2; v <= n; v++) {
        freq[dist[v]]++;
    }

    // Build partial sums: S(i) = freq[1] + freq[2] + ... + freq[i]
    // We only need i up to maxD
    long long runningSum = 0;
    long long answer = 0; // This will store the minimal K
    for (int i = 1; i <= maxD; i++) {
        runningSum += freq[i]; // S(i)
        // We need i * K >= S(i), i.e. K >= S(i)/i
        // So K >= ceil(S(i)/i) = (S(i) + i - 1) // i (integer division)
        long long needed = (runningSum + i - 1) / i;
        answer = max(answer, needed);
    }

    cout << answer << "\n";
    return 0;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10