이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "escape_route.h"
#include <bits/stdc++.h>
#define pb push_back
#define f first
#define sc second
using namespace std;
typedef long long int ll;
typedef string str;
const ll inf = 5e17;
vector<ll> calculate_necessary_time(int n, int m, ll S, int q, vector<int> A, vector<int> B,
vector<ll> L, vector<ll> C, vector<int> U, vector<int> V, vector<ll> T){
vector<ll> ans(q, inf);
vector<vector<tuple<int, ll, ll>>> v(n);
for(int i = 0; i < m; i++){
v[A[i]].pb({B[i], L[i], C[i]});
v[B[i]].pb({A[i], L[i], C[i]});
}
vector<vector<ll>> dis2(n);
for(int s = 0; s < n; s++){
dis2[s].assign(n, inf);
vector<bool> bl(n, 0);
priority_queue<pair<ll, int>> pq;
dis2[s][s] = 0;
pq.push({0, s});
while(!pq.empty()){
int nd = pq.top().sc; pq.pop();
if(bl[nd]) continue;
bl[nd] = 1;
for(auto [x, l, c]: v[nd])
if((dis2[s][nd]%S+l<=c ? dis2[s][nd]+l:((dis2[s][nd]+S-1)/S)*S+l) < dis2[s][x]){
dis2[s][x] = (dis2[s][nd]%S+l<=c ? dis2[s][nd]+l:((dis2[s][nd]+S-1)/S)*S+l);
pq.push({-dis2[s][x], x});
}
}
}
vector<vector<vector<ll>>> dis1;
vector<ll> sth;
ll t = 0;
while(1){
ll nxt = inf;
sth.pb(t);
dis1.pb(vector<vector<ll>>(n));
vector<vector<ll>> &dis = dis1.back();
for(int s = 0; s < n; s++){
dis[s].assign(n, inf);
vector<ll> change(n, inf);
vector<bool> bl(n, 0);
priority_queue<pair<ll, int>> pq;
dis[s][s] = t;
pq.push({0, s});
while(!pq.empty()){
int nd = pq.top().sc; pq.pop();
if(bl[nd]) continue;
bl[nd] = 1;
for(auto [x, l, c]: v[nd]) if(dis[s][nd]+l <= min(dis[s][x]-1, c)){
dis[s][x] = dis[s][nd]+l;
change[x] = c-dis[s][x]+1;
pq.push({-dis[s][x], x});
}
}
for(ll &x: dis[s]) if(x != inf) x-=t;
nxt = min(nxt, *min_element(change.begin(), change.end()));
}
if(nxt == inf) break;
t+=nxt;
}
for(int i = 0; i < q; i++){
int p = (upper_bound(sth.begin(), sth.end(), T[i])-sth.begin())-1;
if(dis1[p][U[i]][V[i]] != inf){
ans[i] = min(ans[i], dis1[p][U[i]][V[i]]);
continue;
}
for(int j = 0; j < n; j++) if(dis1[p][U[i]][j] != inf)
ans[i] = min(ans[i], S-T[i]+dis2[j][V[i]]);
}
return ans;
}
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