이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define ll long long
#define ar array
#define db double
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define rint(l, r) uniform_int_distribution<int>(l, r)(rng)
template<typename T> bool ckmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
void test_case() {
int n, m;
cin >> n >> m;
int S, T, U, V;
cin >> S >> T >> U >> V;
vector<vector<ar<int, 2>>> adj(n+1);
for (int i = 0; i < m; i++) {
int x, y, w;
cin >> x >> y >> w;
adj[x].push_back({y, w});
adj[y].push_back({x, w});
}
auto cost = [&](int sx) -> vector<ll> {
vector<ll> d(n+1, 1e18);
priority_queue<ar<ll, 2>, vector<ar<ll, 2>>, greater<ar<ll, 2>>> q;
q.push({0, sx});
d[sx] = 0;
while (q.size()) {
auto [weight, v] = q.top(); q.pop();
if (weight != d[v]) continue;
for (auto [u, w] : adj[v]) {
if (ckmin(d[u], d[v] + w)) {
q.push({d[u], u});
}
}
}
return d;
};
vector<ll> ds(cost(S)), dt(cost(T)), du(cost(U)), dv(cost(V));
// sort by order of dv
vector<int> each(n);
iota(all(each), 1);
sort(all(each), [&](int x, int y) { return dv[x] < dv[y]; });
vector<int> where(n+1);
for (int i = 0; i < n; i++) where[each[i]] = i;
ll ans = 1e18, bst = ds[T];
vector<int> dp1(n+1, 1e9), dp2(n+1, 1e9);
vector<int> o1(n);
iota(all(o1), 1);
sort(all(o1), [&](int x, int y) { return ds[x] < ds[y]; });
for (int v : o1) { // v --> s
if (ds[v] + dt[v] != bst) continue;
dp1[v] = where[v];
for (auto [u, w] : adj[v]) {
if (dt[v] + w + ds[u] == bst) {
ckmin(dp1[v], dp1[u]);
}
}
}
sort(all(o1), [&](int x, int y) { return dt[x] < dt[y]; });
for (int v : o1) { // v --> t
if (ds[v] + dt[v] != bst) continue;
dp2[v] = where[v];
for (auto [u, w] : adj[v]) {
if (ds[v] + w + dt[u] == bst) {
ckmin(dp2[v], dp2[u]);
}
}
}
for (int i = 1; i <= n; i++) {
int idx = min(dp1[i], dp2[i]);
if (idx < n) {
ckmin(ans, du[i] + dv[each[idx]]);
}
}
cout << min(ans, du[V]) << '\n';
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int t = 1;
// cin >> t;
while (t--) test_case();
}
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