Submission #1139362

#	Time	Username	Problem	Language	Result	Execution time	Memory
1139362		monkey133	Ski 2 (JOI24_ski2)	C++20	100 / 100	362 ms	434732 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<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


ll dp[605][305][305];

ll cal(ll n, ll k, ll gap)
{
	ll g = min((n/k), gap);
	return (g)*(g-1)/2*k + g*(n - k * g);
}

int main()
{
	ll n, ck;
	cin >> n >> ck;
	vector<ll> dis;
	vector<pi> a(n);
	for (ll i=0; i<n; ++i) {cin >> a[i].f >> a[i].s; dis.pb(a[i].f); dis.pb(a[i].f+1);}
	discretize(dis);
	ll nn = dis.size();
	vector<ll> min_cost(nn, ll(1e15)), cnt(nn, 0);
	for (ll i=0; i<n; ++i)
	{
		a[i].f = lb(all(dis), a[i].f)-dis.begin();
		cnt[a[i].f]++;
		min_cost[a[i].f] = min(min_cost[a[i].f], a[i].s);
	}
	for (ll i=0; i<=nn; ++i)for (ll j=0; j<=n; ++j)for (ll k=0; k<=n; ++k) dp[i][j][k] = ll(1e15);
	dp[0][1][0] = 0;
	ll ps = ll(1e15);
	for (ll i=0; i<nn; ++i)
	{
		ll gap = ((i==nn-1)?n:dis[i+1] - dis[i]);
		for (ll j=1; j<=n; ++j)
		{
			for (ll k=0; k<=n; ++k)
			{
				// buy more connection
				if (j + 1 <= n) dp[i][j + 1][k] = min(dp[i][j + 1][k], dp[i][j][k] + ps);
				// just connect to the next one
				dp[i+1][j][max(0LL, cnt[i] + k - j*gap)] = min(dp[i+1][j][max(0LL, cnt[i] + k - j*gap)], dp[i][j][k] + ck * cal(k + cnt[i], j, gap));
				//show(dp[i][j][k]);
			}
		}
		ps = min(ps, min_cost[i]);
	}
	ll ans = ll(1e15);
	for (ll i=1; i<=n; ++i) { ans = min(ans, dp[nn][i][0]);}
	cout << ans << endl;
	
	return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 12 / 12
• Subtask 3: 9 / 9
• Subtask 4: 33 / 33
• Subtask 5: 27 / 27
• Subtask 6: 14 / 14