Submission #985355

#	Time	Username	Problem	Language	Result	Execution time	Memory
985355		CrazyBotBoy	Cyberland (APIO23_cyberland)	C++17	15 / 100	835 ms	2097152 KiB

cyberland

#include <iostream>
#include <vector>
#include <queue>
#include <limits>
#include <iomanip>

const double INF = std::numeric_limits<double>::infinity();

struct State {
    double time;
    int country;
    int div2_used;

    bool operator>(const State& other) const {
        return time > other.time;
    }
};

double solve(int N, int M, int K, int H, std::vector<int> x, std::vector<int> y, std::vector<int> c, std::vector<int> arr) {
    std::vector<std::vector<std::pair<int, int>>> graph(N);

    // Build the graph
    for (int i = 0; i < M; ++i) {
        graph[x[i]].emplace_back(y[i], c[i]);
        graph[y[i]].emplace_back(x[i], c[i]);
    }

    // Priority queue for Dijkstra-like algorithm
    std::priority_queue<State, std::vector<State>, std::greater<State>> pq;
    pq.push({0, 0, 0});

    // Distance table to store the minimum time to reach each country with a certain number of divide-by-2 used
    std::vector<std::vector<double>> dist(N, std::vector<double>(K + 1, INF));
    dist[0][0] = 0;

    while (!pq.empty()) {
        auto [current_time, current_country, div2_used] = pq.top();
        pq.pop();

        if (current_country == H) {
            return current_time;
        }

        // Early exit if we already have found a better way
        if (current_time > dist[current_country][div2_used]) {
            continue;
        }

        // Explore all adjacent countries
        for (const auto& [next_country, travel_time] : graph[current_country]) {
            double next_time = current_time + travel_time;

            // Case 1: Without using any special ability
            if (next_time < dist[next_country][div2_used]) {
                dist[next_country][div2_used] = next_time;
                pq.push({next_time, next_country, div2_used});
            }

            // Case 2: Using special abilities
            if (arr[next_country] == 0) {
                next_time = 0;
                if (next_time < dist[next_country][div2_used]) {
                    dist[next_country][div2_used] = next_time;
                    pq.push({next_time, next_country, div2_used});
                }
            } else if (arr[next_country] == 2 && div2_used < K) {
                next_time = current_time / 2.0;
                if (next_time < dist[next_country][div2_used + 1]) {
                    dist[next_country][div2_used + 1] = next_time;
                    pq.push({next_time, next_country, div2_used + 1});
                }
            }
        }
    }

    return -1;  // If we exhaust the priority queue and don't find a way to H
}

• Subtask 1: 0 / 5
• Subtask 2: 8 / 8
• Subtask 3: 0 / 13
• Subtask 4: 0 / 19
• Subtask 5: 7 / 7
• Subtask 6: 0 / 16
• Subtask 7: 0 / 29
• Subtask 8: 0 / 3