Submission #985152

#TimeUsernameProblemLanguageResultExecution timeMemory
985152Kracken_180Cyberland (APIO23_cyberland)C++17
20 / 100
2127 ms2097152 KiB
#include <iostream> #include <vector> #include <queue> #include <tuple> #include <limits> using namespace std; double solve(int N, int M, int K, int H, vector<int> x, vector<int> y, vector<int> c, vector<int> arr) { const double INF = numeric_limits<double>::infinity(); // Adjacency list representation of the graph vector<vector<pair<int, int>>> graph(N); 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 to store the states as (current cost, current country, remaining K) priority_queue<tuple<double, int, int>, vector<tuple<double, int, int>>, greater<>> pq; pq.emplace(0.0, 0, K); // Distance table to store the minimum cost to reach each state vector<vector<double>> dist(N, vector<double>(K + 1, INF)); dist[0][K] = 0.0; while (!pq.empty()) { auto [current_cost, u, remaining_k] = pq.top(); pq.pop(); if (u == H) { return current_cost; } // If we find a cheaper way to get here, continue if (current_cost > dist[u][remaining_k]) { continue; } for (auto &[v, travel_time] : graph[u]) { double new_cost = current_cost + travel_time; // Case 1: Normal travel without using special abilities if (new_cost < dist[v][remaining_k]) { dist[v][remaining_k] = new_cost; pq.emplace(new_cost, v, remaining_k); } // Case 2: Using zeroing ability if (arr[v] == 0 && current_cost < dist[v][remaining_k]) { dist[v][remaining_k] = current_cost; pq.emplace(current_cost, v, remaining_k); } // Case 3: Using halving ability if (arr[v] == 2 && remaining_k > 0) { double halved_cost = current_cost + travel_time / 2.0; if (halved_cost < dist[v][remaining_k - 1]) { dist[v][remaining_k - 1] = halved_cost; pq.emplace(halved_cost, v, remaining_k - 1); } } } } // If Cyberland is unreachable return -1.0; }
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