Submission #1272841

#	Time	Username	Problem	Language	Result	Execution time	Memory
1272841		soab	Airplane (NOI23_airplane)	C++20	0 / 100	1 ms	568 KiB

Main

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

const long long INF = 1e18; // Use a large value for infinity

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

    int n;
    int m;
    cin >> n >> m;

    vector<long long> a(n + 1);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }

    vector<vector<int>> adj(n + 1);
    for (int i = 0; i < m; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    // --- Dijkstra 1: Calculate up[i] from source 1 ---
    vector<long long> up(n + 1, INF);
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq_up;

    up[1] = a[1]; // Start time at region 1 is its altitude requirement (which is 0)
    pq_up.push({up[1], 1});

    while (!pq_up.empty()) {
        long long time_u = pq_up.top().first;
        int u = pq_up.top().second;
        pq_up.pop();

        if (time_u > up[u]) {
            continue;
        }

        for (int v : adj[u]) {
            long long new_time_v = max(time_u + 1, a[v]);
            if (new_time_v < up[v]) {
                up[v] = new_time_v;
                pq_up.push({up[v], v});
            }
        }
    }

    // --- Dijkstra 2: Calculate down[i] from source n ---
    vector<long long> down(n + 1, INF);
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq_down;

    down[n] = a[n]; // Start time at region n is its altitude requirement (which is 0)
    pq_down.push({down[n], n});

    while (!pq_down.empty()) {
        long long time_v = pq_down.top().first;
        int v = pq_down.top().second;
        pq_down.pop();

        if (time_v > down[v]) {
            continue;
        }

        for (int u : adj[v]) { // Iterating neighbors
            long long new_time_u = max(time_v + 1, a[u]);
            if (new_time_u < down[u]) {
                down[u] = new_time_u;
                pq_down.push({down[u], u});
            }
        }
    }

    // --- Combine results to find the minimum total time ---
    long long min_total_time = INF;
    for (int i = 1; i <= n; ++i) {
        min_total_time = min(min_total_time, up[i] + down[i]);
    }

    cout << min_total_time << endl;

    return 0;
}

• Subtask 1: 0 / 22
• Subtask 2: 0 / 10
• Subtask 3: 0 / 31
• Subtask 4: 0 / 37