Submission #1166333

#	Time	Username	Problem	Language	Result	Execution time	Memory
1166333		merciless_lassie	Commuter Pass (JOI18_commuter_pass)	C++20	0 / 100	247 ms	327680 KiB

commuter_pass

#include "bits/stdc++.h"
using namespace std;
#define int long long
const int INF = 1e18;

int n, m;
int s, t, u, v;
vector<pair<int, int>> graph[100005];
vector<vector<int>> shortest_path_edges;

// Find shortest path from src to all nodes
vector<int> dijkstra(int src) {
    vector<int> dist(n + 1, INF);
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> 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 edge : graph[node]) {
            int next = edge.first;
            int weight = edge.second;
            
            if (dist[node] + weight < dist[next]) {
                dist[next] = dist[node] + weight;
                pq.push({dist[next], next});
            }
        }
    }
    
    return dist;
}

// Mark all edges that could be part of any shortest path from S to T
void markShortestPathEdges(vector<int>& distFromS, vector<int>& distToT) {
    shortest_path_edges.resize(n + 1, vector<int>(n + 1, 0));
    
    for (int i = 1; i <= n; i++) {
        for (auto edge : graph[i]) {
            int j = edge.first;
            int weight = edge.second;
            
            // If this edge is part of any shortest path from S to T
            if (distFromS[i] + weight + distToT[j] == distFromS[t] || 
                distFromS[j] + weight + distToT[i] == distFromS[t]) {
                shortest_path_edges[i][j] = 1;
                shortest_path_edges[j][i] = 1;
            }
        }
    }
}

// Find minimum cost from U to V with commuter pass
int findMinCostUtoV(vector<int>& distFromS, vector<int>& distToT) {
    // Build a new graph where edges in shortest path are free
    vector<pair<int, int>> new_graph[100005];
    
    for (int i = 1; i <= n; i++) {
        for (auto edge : graph[i]) {
            int j = edge.first;
            int weight = edge.second;
            
            if (shortest_path_edges[i][j]) {
                // Edge is part of shortest path, so it's free
                new_graph[i].push_back({j, 0});
            } else {
                // Edge is not part of shortest path
                new_graph[i].push_back({j, weight});
            }
        }
    }
    
    // Run Dijkstra from U to find minimum cost to V
    vector<int> dist(n + 1, INF);
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> 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 edge : new_graph[node]) {
            int next = edge.first;
            int weight = edge.second;
            
            if (dist[node] + weight < dist[next]) {
                dist[next] = dist[node] + weight;
                pq.push({dist[next], next});
            }
        }
    }
    
    return dist[v];
}

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});
    }
    
    // Find shortest paths
    vector<int> distFromS = dijkstra(s);
    vector<int> distToT(n + 1, INF);
    
    // Calculate distance to T by running Dijkstra from T
    vector<int> distFromT = dijkstra(t);
    for (int i = 1; i <= n; i++) {
        distToT[i] = distFromT[i];
    }
    
    // Mark all edges that can be part of any shortest path from S to T
    markShortestPathEdges(distFromS, distToT);
    
    // Find minimum cost from U to V
    int result = findMinCostUtoV(distFromS, distToT);
    
    cout << result << endl;
    
    return 0;
}

• Subtask 1: 0 / 16
• Subtask 2: 0 / 15
• Subtask 3: 0 / 24
• Subtask 4: 0 / 45