Submission #1241956

#	Time	Username	Problem	Language	Result	Execution time	Memory
1241956		khanhtt	Commuter Pass (JOI18_commuter_pass)	C++20	16 / 100	219 ms	24504 KiB

commuter_pass

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = (ll)1e18;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M, S, T, U, V;
    cin >> N >> M >> S >> T >> U >> V;

    vector<vector<pair<int,int>>> adj(N+1);
    for(int i = 0; i < M; i++){
        int A, B, C;
        cin >> A >> B >> C;
        adj[A].emplace_back(B,C);
        adj[B].emplace_back(A,C);
    }

    auto dijkstra = [&](int src){
        vector<ll> dist(N+1, INF);
        priority_queue<pair<ll,int>,
                       vector<pair<ll,int>>,
                       greater<>> pq;
        dist[src] = 0;
        pq.push({0, src});
        while(!pq.empty()){
            auto [d, x] = pq.top(); pq.pop();
            if(d > dist[x]) continue;
            for(auto [y, w] : adj[x]){
                if(d + w < dist[y]){
                    dist[y] = d + w;
                    pq.push({dist[y], y});
                }
            }
        }
        return dist;
    };

    // 1) three Dijkstra’s
    auto du = dijkstra(U);
    auto dv = dijkstra(V);
    auto ds = dijkstra(S);
    auto dt = dijkstra(T);

    ll D = ds[T];
    if(D == INF){
        cout << "-1\n";
        return 0;
    }

    // 2) Build the shortest-path DAG from S
    vector<vector<int>> dag(N+1), rdag(N+1);
    for(int x = 1; x <= N; x++){
        for(auto [y, w] : adj[x]){
            if(ds[x] + w == ds[y]){
                dag[x].push_back(y);
                rdag[y].push_back(x);
            }
        }
    }

    // 3) Forward DP on dag to get F[i] = min dv[] on some S→i prefix
    vector<ll> F(N+1, INF);
    F[S] = dv[S];
    // nodes sorted by increasing ds[]
    vector<int> order(N);
    iota(order.begin(), order.end(), 1);
    sort(order.begin(), order.end(),
         [&](int a, int b){ return ds[a] < ds[b]; });
    for(int x : order){
        if(F[x] == INF) continue;
        for(int y : dag[x]){
            ll cand = min(F[x], dv[y]);
            if(cand < F[y]) F[y] = cand;
        }
    }

    // 4) Backward DP on rdag to get G[i] = min dv[] on some i→T suffix
    vector<ll> G(N+1, INF);
    G[T] = dv[T];
    // same order, but decreasing ds[]
    sort(order.begin(), order.end(),
         [&](int a, int b){ return ds[a] > ds[b]; });
    for(int x : order){
        if(G[x] == INF) continue;
        for(int p : rdag[x]){
            ll cand = min(G[x], dv[p]);
            if(cand < G[p]) G[p] = cand;
        }
    }

    // 5) Over all i on some shortest S→T path (ds[i]+dt[i] == D),
    //    minimize du[i] + min(F[i], G[i])
    ll ans = INF;
    for(int i = 1; i <= N; i++){
        if(ds[i] + dt[i] == D){
            // i lies on at least one shortest S→T
            ll bestDvOnPath = min(F[i], G[i]);
            if(bestDvOnPath < INF && du[i] < INF){
                ans = min(ans, du[i] + bestDvOnPath);
            }
        }
    }

    cout << ans << "\n";
    return 0;
}

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