Submission #922102

#	Time	Username	Problem	Language	Result	Execution time	Memory
922102		406	Commuter Pass (JOI18_commuter_pass)	C++17	31 / 100	362 ms	29164 KiB

commuter_pass

#include <bits/stdc++.h>
#define int int64_t
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)

using namespace std;
using ar = array<int, 2>;

const int64_t INF = 1ll << 60;
const int N = 2e5 + 5;

int n, m, S, T, U, V, d[4][N], mnd;
vector<ar> adj[N];
bool mark[N], mark2[N];

void dfs(int v) {
        mark[v] = true;
        for (auto [u, w]: adj[v]) if (!w && !mark[u])
                dfs(u);
        mnd = min(mnd, d[2][v]);
}
void dfs2(int v) {
        mark2[v] = true;
        for (auto [u, w]: adj[v]) if (!w && !mark2[u])
                dfs2(u);
        d[2][v] = min(d[2][v], mnd);
}

void dij(int x, int d[]) {
        fill(d, d + N, INF);
        priority_queue<ar> q;
        q.push({d[x] = 0, x});
        while (q.size()) {
                auto [di, v] = q.top();
                q.pop();
                if (-di > d[v]) continue;
                for (auto [u, w]: adj[v]) {
                        w += d[v];
                        if (w < d[u]) {
                                d[u] = w;
                                q.push({-d[u], u});
                        }
                }
        }
}

signed main() {
        ios::sync_with_stdio(false);
        cin.tie(nullptr); 
        cin >> n >> m >> S >> T >> U >> V;
        --S, --T, --U, --V;
        FOR(i, 0, m) {
                int u, v, w;
                cin >> u >> v >> w;
                --u, --v;
                adj[v].push_back({u, w});
                adj[u].push_back({v, w});
        }
        dij(T, d[0]);
        dij(S, d[1]);
        dij(U, d[2]);
        dij(V, d[3]);

        FOR(v, 0, n) for (auto &[u, w]: adj[v]) 
                w = w + d[0][v] + d[1][u] == d[0][S] ? 0 : w;
        /*
        FOR(i, 0, n) if (!mark[i]) {
                mnd = INF;
                dfs(i);
                dfs2(i);
        }
        int mn = INF;
        FOR(i, 0, n) mn = min(mn, d[2][i] + d[3][i]);
        */
        dij(U, d[2]);
        dij(V, d[3]);
        cout << min(d[2][V], d[3][U]) << '\n';
        return 0;
}

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