Submission #957287

#TimeUsernameProblemLanguageResultExecution timeMemory
957287amine_arouaToll (BOI17_toll)C++17
100 / 100
84 ms15708 KiB
#include <bits/stdc++.h> //#pragma GCC optimize("O3") //#pragma GCC optimize("unroll-loops") using namespace std; #define int long long #define vi vector<int> #define vl vector<long long> #define vii vector<pair<int,int>> #define vll vector<pair<long long,long long>> #define pb push_back #define ll long long #define ld long double #define nl '\n' #define boost ios::sync_with_stdio(false) #define mp make_pair #define se second #define fi first #define fore(i, y) for(int i = 0; i < y; i++) #define forr(i,x,y) for(int i = x;i<=y;i++) #define forn(i,y,x) for(int i = y; i >= x; i--) #define all(v) v.begin(),v.end() #define sz(v) (int)v.size() #define clr(v,k) memset(v,k,sizeof(v)) #define rall(v) v.rbegin() , v.rend() #define pii pair<int,int> #define pll pair<ll , ll> #define vvi vector<vi> #define vvii vector<vii> const ll MOD = 1e9 + 7; const ll INF = 1e18 + 1; ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (gcd) ll lcm(ll a , ll b) {return a * (b / gcd(a , b));} // least common multiple (lcm) // HERE IS THE SOLUTION int N , K , M , O; const int LOG = 15; int lim; vvii adj; vector<vi> comb(vector<vi> a , vector<vi> b) { vector<vi> target(K , vi(K , INF)); fore(x , K) { fore(y , K) { fore(z , K) { target[x][y] = min(target[x][y] , a[x][z] + b[z][y]); } } } return target; } vector<vvi> tree; vvi query(int node , int L , int R , int l , int r) { if(l <= L && R <= r) return tree[node]; int mid = (L + R)/2; if(r <= mid) return query(2*node , L , mid , l , r); if(l > mid) return query(2*node + 1, mid + 1, R , l, r); return comb(query(2*node , L , mid , l , r) , query(2*node + 1 , mid + 1 , R , l , r)); } signed main() { boost; cin.tie(0); cout.tie(0); cin>>K>>N>>M>>O; lim = (N+ K - 1)/K; while(__builtin_popcount(lim) != 1) { lim++; } tree.assign(2*lim , vvi(K , vi(K , INF))); fore(i , M) { int a , b ,t; cin>>a>>b>>t; tree[(a/K) + lim][a%K][b%K] = min(tree[(a/K) + lim][a%K][b%K] , t); } forn(i , lim - 1 ,1) tree[i] = comb(tree[2*i] , tree[2*i + 1]); while(O--) { int a , b; cin>>a>>b; if((a/K) >= (b/K)) { cout<<-1<<nl; continue; } int ans = query(1 , 0 , lim - 1, a/K , (b/K) - 1)[a%K][b%K]; if(ans >= INF) cout<<-1<<nl; else cout<<ans<<nl; } }
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