제출 #41411

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
41411		RockyB	Jakarta Skyscrapers (APIO15_skyscraper)	C++14	10 / 100	22 ms	5276 KiB

skyscraper

/// In The Name Of God
 
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#include <bits/stdc++.h>
 
#define f first
#define s second
 
#define pb push_back
#define pp pop_back
#define mp make_pair
 
#define sz(x) (int)x.size()
#define sqr(x) ((x) * 1ll * (x))
#define all(x) x.begin(), x.end()
 
#define Kazakhstan ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);
 
#define nl '\n'
#define ioi exit(0);
 
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
 
const int N = (int)3e4 + 7;
const int inf = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const ll linf = (ll)1e18 + 7;
 
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
 
using namespace std;
 
int n, m;
int sta, fn;
int b[N], p[N];
vector < pair <int, int> > g[N];
vector <int> a[N], dv[N];

ll d[N];
int main() {
	#ifdef IOI2018
		freopen ("in.txt", "r", stdin);
	#endif
	Kazakhstan
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		cin >> b[i] >> p[i];
		a[b[i]].pb(p[i]);
		if (i == 1) sta = b[i];
		if (i == 2) fn = b[i];
	}
	for (int i = 1; i <= n; i++) {
		if (sz(a[i])) sort(all(a[i]));
	}

	for (int i = 30000; i >= 1; i--) {
		for (int j = i; j <= 30000; j += i) {
			dv[j].pb(i);
		}
	}

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (i == j) continue;
			int df = abs(i - j);
			for (auto it : dv[df]) {
				if (binary_search(all(a[i]), it)) {
					g[i].pb({j, df / it});
					break;
				}
			}
		}
	}
	
	set < pair <ll, int > > st;
	memset(d, 0x3f, sizeof(d));
	st.insert({0, sta});
	d[sta] = 0;
	while (sz(st)) {
		pair <ll, int> v = *st.begin();
		st.erase(st.begin());
		for (auto to : g[v.s]) {
			if (d[to.f] > v.f + to.s) {
				st.erase({d[to.f], to.f});
				d[to.f] = v.f + to.s;
				st.insert({d[to.f], to.f});
			}
		}
	}
	if (d[fn] >= linf) cout << -1;
	else cout << d[fn];
	ioi
}

• Subtask 1: 10 / 10
• Subtask 2: 0 / 12
• Subtask 3: 0 / 14
• Subtask 4: 0 / 21
• Subtask 5: 0 / 43