이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//#pragma GCC optimize("O3,unroll-loops,inline,fast-math")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define all(a) (a).begin(), (a).end()
#define int ll
constexpr int INF = numeric_limits<int>::max() / 2;
vector<int> dijkstra(vector<vector<pair<int, int>>> &gr, int s) {
int n = gr.size();
vector<int> dist(n, INF);
dist[s] = 0;
set<pair<int, int>> st;
st.insert({0, s});
while (!st.empty()) {
int v = st.begin()->second;
st.erase(st.begin());
for (auto [u, cost] : gr[v])
if (dist[u] > dist[v] + cost) {
st.erase({dist[u], u});
dist[u] = dist[v] + cost;
st.insert({dist[u], u});
}
}
return dist;
}
void solve() {
int n, m, s, t, u, v;
cin >> n >> m >> s >> t >> u >> v;
--s, --t, --u, --v;
vector<vector<pair<int, int>>> gr(n);
for (int i = 0, a, b, c; i < m; ++i) {
cin >> a >> b >> c;
--a, --b;
gr[a].push_back({b, c});
gr[b].push_back({a, c});
}
auto ds = dijkstra(gr, s), dt = dijkstra(gr, t);
vector<vector<int>> dag(n);
for (int i = 0; i < n; ++i)
for (auto [j, cost] : gr[i])
if (ds[i] + dt[j] + cost == ds[t])
dag[i].push_back(j);
auto solve = [&](vector<int> &du, vector<int> dv) -> int {
int res = INF;
vector<bool> used(n, false);
auto dfs = [&](auto dfs, int a) -> int {
if (!used[a]) {
used[a] = true;
for (int b : dag[a])
dv[a] = min(dv[a], dfs(dfs, b));
res = min(res, dv[a] + du[a]);
}
return dv[a];
};
for (int i = 0; i < n; ++i)
dfs(dfs, i);
return res;
};
auto du = dijkstra(gr, u), dv = dijkstra(gr, v);
cout << min(solve(du, dv), solve(dv, du));
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t = 1;
// cin >> t;
while (t--)
solve();
}
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