제출 #1159967

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1159967		pan	Akcija (COCI21_akcija)	C++20	110 / 110	1426 ms	63516 KiB

Main

#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<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll n, k;
pi dp[2005][2005];
vector<pi> v, ans;
pi dfs(ll idx, ll cnt)
{
	if (idx == n) return mp(0, 0);
	if (dp[idx][cnt] != mp(-1LL, -1LL)) return dp[idx][cnt];
	pi ans = dfs(idx + 1, cnt);
	if (cnt < v[idx].f) 
	{
		pi ans2 = dfs(idx + 1, cnt + 1);
		ans2.f++; ans2.s+= v[idx].s;
		ans = max(ans, ans2);
	}
	return dp[idx][cnt] = ans;
}

void dfs2(ll idx, ll cnt, ll c, ll w)
{
	if (ans.size() == k) return;
	if (idx==n)
	{
		if (mp(c, w) < mp(0LL, 0LL)) ans.pb(mp(c, w));
		return;
	}
	if (mp(c, w) >= dfs(idx, cnt)) return; // no solution found
	dfs2(idx + 1, cnt, c, w);
	if (ans.size() == k) return;
	if (cnt < v[idx].f) dfs2(idx + 1, cnt  + 1, c-1, w - v[idx].s);
}

int main()
{
	cin >> n >> k;
	v.resize(n);
	for (ll i=0; i<=n; ++i) for (ll j=0; j<=n; ++j) dp[i][j] = mp(-1, -1);
	for (ll i=0; i<n; ++i) cin >> v[i].s >> v[i].f;
	sort(all(v));
	for (ll i=0; i<n; ++i) {v[i].s*=-1; v[i].s += 1e9;}
	ll lo = 0, hi = 2e16, cut = 2e9*2000;
	while (lo!=hi)
	{
		ll mid = (lo + hi + 1) >> 1;
		ans.clear();
		dfs2(0, 0, mid/cut, mid%cut);
		if (ans.size() < k) hi = mid-1;
		else lo = mid;
	}
	//show(lo);
	ans.clear();
	dfs2(0, 0, lo/cut, lo%cut);
	for (ll i=0; i<ans.size(); ++i) ans[i].f = -ans[i].f + lo/cut, ans[i].s = (-ans[i].s + lo%cut);
	sort(all(ans), greater<pi> ());
	while (ans.size() < k) ans.pb(mp(lo/cut, lo%cut));
	for (ll i=0; i<k; ++i) cout << ans[i].f << ' ' << -ans[i].s  + ll(1e9*ans[i].f) << endl;
	return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 10 / 10
• Subtask 5: 30 / 30
• Subtask 6: 20 / 20