Submission #1166367

#TimeUsernameProblemLanguageResultExecution timeMemory
1166367merciless_lassieCommuter Pass (JOI18_commuter_pass)C++20
31 / 100
199 ms18584 KiB
#include "bits/stdc++.h" using namespace std; #define int long long const int INF = 1e18; int n, m, s, t, u, v; vector<pair<int, int>> graph[100005]; // Standard Dijkstra to compute shortest distances from a source vector<int> dijkstra(int src) { vector<int> dist(n + 1, INF); priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> pq; dist[src] = 0; pq.push({0, src}); while (!pq.empty()) { int d = pq.top().first; int node = pq.top().second; pq.pop(); if (d > dist[node]) continue; for (auto [next, weight] : graph[node]) { if (dist[node] + weight < dist[next]) { dist[next] = dist[node] + weight; pq.push({dist[next], next}); } } } return dist; } // Modified Dijkstra for U to V considering S-T shortest paths int dijkstra_uv_with_pass(vector<int>& dist_s, vector<int>& dist_t) { vector<int> dist(n + 1, INF); priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> pq; dist[u] = 0; pq.push({0, u}); while (!pq.empty()) { int d = pq.top().first; int node = pq.top().second; pq.pop(); if (d > dist[node]) continue; for (auto [next, weight] : graph[node]) { // Check if this edge can be on some S-T shortest path bool is_free = (dist_s[node] + weight + dist_t[next] == dist_s[t]) || (dist_s[next] + weight + dist_t[node] == dist_s[t]); int cost = is_free ? 0 : weight; if (dist[node] + cost < dist[next]) { dist[next] = dist[node] + cost; pq.push({dist[next], next}); } } } return dist[v]; } int solve() { // Compute shortest distances from S and T vector<int> dist_s = dijkstra(s); vector<int> dist_t = dijkstra(t); if (dist_s[t] == INF) return -1; // No path from S to T (shouldn't happen per constraints) // Compute minimum cost from U to V considering the commuter pass return dijkstra_uv_with_pass(dist_s, dist_t); } signed main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); cin >> n >> m; cin >> s >> t >> u >> v; for (int i = 0; i < m; i++) { int a, b, c; cin >> a >> b >> c; graph[a].push_back({b, c}); graph[b].push_back({a, c}); } cout << solve() << endl; return 0; }
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