Submission #385816

#	Time	Username	Problem	Language	Result	Execution time	Memory
385816		idontreallyknow	Semiexpress (JOI17_semiexpress)	C++17	100 / 100	9 ms	5484 KiB

semiexpress

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
template <class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class A> ostream& operator<<(ostream &cout, vector<A> const &v) {cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";};
template<class A, class B> ostream& operator<<(ostream &cout, const pair<A,B> &x) {return cout << "(" <<x.first << ", " << x.second << ")";};
template <class T> void pv(T a, T b) {cerr << "["; for (T i = a; i != b; ++i) cerr << *i << " "; cerr << "]\n";}
void _print() {cerr << "]\n";}
template <class T, class... V> void _print(T t, V... v) {cerr << t; if (sizeof...(v)) cerr << ", "; _print(v...);}
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#define fi first
#define se second
#define SZ(x) (int)((x).size())
#define pii pair<ll,ll>
const int mx = 3e3+5;
ll n,m,k,a,b,c,t,s[mx];
vector<ll> amt[mx];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m >> k >> a >> b >> c >> t;
	k -= m;
	ll ans = 0;
	for (int q = 1; q <= m; q++) cin >> s[q];
	for (int q = 1; q <= m-1; q++) {
		ll rem = t-(s[q]-s[1])*b;
		if (rem < 0) break;
		ll cur = s[q];
		for (int w = 0; w <= k; w++) {
			if (cur >= s[q+1] || rem < 0) break;
			ll z = rem/a;
			ll en = min(cur+z,s[q+1]-1);
			if (w == 0) ans += (en-cur+1);
			else amt[q].push_back(en-cur+1);
			rem -= (en-cur+1)*c;
			cur = en+1;
		}
	}
	for (int q = 1; q <= m-1; q++) reverse(amt[q].begin(), amt[q].end());
	priority_queue<pii> pq;
	for (int q = 1; q <= m-1; q++) {
		if (SZ(amt[q])) pq.push({amt[q].back(),q});
	}
	while (k > 0 && SZ(pq)) {
		pii x = pq.top();
		pq.pop();
		ans += x.fi;
		amt[x.se].pop_back();
		if (SZ(amt[x.se])) pq.push({amt[x.se].back(),x.se});
		k--;
	}
	if ((s[m]-s[1])*b <= t) ans++;
	cout << ans-1 << "\n";
	return 0;
}

• Subtask 1: 18 / 18
• Subtask 2: 30 / 30
• Subtask 3: 52 / 52