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>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
int N,M,S,T,U,V;
template<int SZ> struct Dijkstra {
ll dist[SZ];
vpi adj[SZ];
priority_queue<pl,vpl,greater<pl>> q;
void addEdge(int A, int B, int C) {
adj[A].pb({B,C}), adj[B].pb({A,C});
}
void gen(int st) {
fill_n(dist,SZ,INF);
q = priority_queue<pl,vpl,greater<pl>>();
dist[st] = 0; q.push({0,st});
while (sz(q)) {
pl x = q.top(); q.pop();
if (dist[x.s] < x.f) continue;
for (pi y: adj[x.s]) if (x.f+y.s < dist[y.f]) {
dist[y.f] = x.f+y.s;
q.push({dist[y.f],y.f});
}
}
}
};
Dijkstra<MX> D;
ll dist[2][MX], DIST[2][MX];
pl mn[MX];
bool ok[MX];
pl comb(pl a, pl b) {
return {min(a.f,b.f),min(a.s,b.s)};
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> M >> S >> T >> U >> V;
F0R(i,M) {
int A,B,C; cin >> A >> B >> C;
D.addEdge(A,B,C);
}
D.gen(U);
FOR(i,1,N+1) dist[0][i] = D.dist[i];
D.gen(V);
FOR(i,1,N+1) dist[1][i] = D.dist[i];
FOR(i,1,N+1) mn[i] = {INF,INF};
D.gen(S);
FOR(i,1,N+1) DIST[0][i] = D.dist[i];
D.gen(T);
FOR(i,1,N+1) DIST[1][i] = D.dist[i];
ll ans = dist[0][V];
vpl v;
FOR(i,1,N+1) if (DIST[0][i]+DIST[1][i] == DIST[0][T]) {
ok[i] = 1;
v.pb({DIST[0][i],i});
}
sort(v.rbegin(),v.rend());
for (auto a: v) {
for (auto t: D.adj[a.s]) if (ok[t.f]) mn[a.s] = comb(mn[a.s],mn[t.f]);
ans = min(ans,min(mn[a.s].f+dist[1][a.s],mn[a.s].s+dist[0][a.s]));
mn[a.s] = comb(mn[a.s],{dist[0][a.s],dist[1][a.s]});
}
cout << ans;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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