Submission #675705

#	Time	Username	Problem	Language	Result	Execution time	Memory
675705		vovamr	Jakarta Skyscrapers (APIO15_skyscraper)	C++17	100 / 100	822 ms	3708 KiB

skyscraper

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define fi first
#define se second
#define ll long long
#define ld long double
#define sz(x) ((int)(x).size())
#define all(x) 	(x).begin(), (x).end()
#define pb push_back
#define mpp make_pair
#define ve vector
using namespace std;
using namespace __gnu_pbds;
template<class T> using oset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
const ll inf = 1e18; const int iinf = 1e9;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T> inline bool chmin(T& a, T b) { return (a > b ? a = b, 1 : 0); }
template <typename T> inline bool chmax(T& a, T b) { return (a < b ? a = b, 1 : 0); }

inline void solve() {
	int n, m;
	cin >> n >> m;

	int fir = -1, sec = -1;

	ve<ve<int>> lst(n);
	for (int i = 0; i < m; ++i) {
		int b, p;
		cin >> b >> p;
		lst[b].pb(p);

		if (!i) fir = b;
		else if (i == 1) sec = b;
	}

	ve<int> d(n, iinf);
	d[fir] = 0;

	set<pii> st;
	st.insert({0, fir});

	while (st.size()) {
		int v = st.begin()->se;
		st.erase(st.begin());

		for (auto &p : lst[v]) {
			int pos, cur;

			pos = v + p, cur = 1;
			while (pos < n) {
				if (d[pos] > d[v] + cur) {
					st.erase({d[pos], pos});
					d[pos] = d[v] + cur;
					st.insert({d[pos], pos});
				}
				pos += p;
				++cur;
			}

			pos = v - p, cur = 1;
			while (pos >= 0) {
				if (d[pos] > d[v] + cur) {
					st.erase({d[pos], pos});
					d[pos] = d[v] + cur;
					st.insert({d[pos], pos});
				}
				pos -= p;
				++cur;
			}
		}
	}
	cout << (d[sec] == iinf ? -1 : d[sec]);
}

signed main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int q = 1; // cin >> q;
	while (q--) solve();
	cerr << fixed << setprecision(3) << "Time execution: " << (double)clock() / CLOCKS_PER_SEC << endl;
}

• Subtask 1: 10 / 10
• Subtask 2: 12 / 12
• Subtask 3: 14 / 14
• Subtask 4: 21 / 21
• Subtask 5: 43 / 43