Submission #1147842

#	Time	Username	Problem	Language	Result	Execution time	Memory
1147842		monkey133	Semiexpress (JOI17_semiexpress)	C++20	100 / 100	1 ms	328 KiB

semiexpress

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll n, m, k, a, b, c, t;
ll p[3005], now[3005];

priority_queue<pii> pq;

void update(ll x)
{
	if ((p[x]-1)*b + (now[x]-p[x])*c > t || now[x] >= p[x + 1]) return;
	//show(x);
	//show((p[x]-1)*b + (now[x]-p[x])*c);
	ll r = min(p[x + 1]-1, now[x] + (t - (p[x]-1)*b - (now[x]-p[x])*c)/a);
	pq.push(mp(r - now[x] + 1, mp(r, x)));
}


int main()
{
	ll ans = 0;
	cin >> n >> m >> k >> a >> b >> c >> t;
	for (ll i=1; i<=m; ++i) cin >> p[i];
	p[m + 1] = n + 1;
	for (ll i=1; i<=m; ++i)
	{
		if ((p[i] -1)*b > t) break;
		//show((t - (p[i]-1)*b)/a);
		ll r = min(p[i + 1]-1, p[i] + (t - (p[i]-1)*b)/a );
		ans += r - p[i] + 1;
		now[i] = r + 1;
		update(i);
		//show2(i, r);
	}
	//show(ans);
	k -= m;
	while (k && pq.size())
	{
		pii u = pq.top();
		//show3(u.f, u.s.f, u.s.s);
		pq.pop();
		ans += u.f;
		now[u.s.s] = u.s.f  + 1;
		update(u.s.s);
		k--;
	}
	cout << ans-1 << endl;
	return 0;
}

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