Submission #1082057

#	Time	Username	Problem	Language	Result	Execution time	Memory
1082057		GrindMachine	Jelly Flavours (IOI20_jelly)	C++17	100 / 100	147 ms	972 KiB

jelly

#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;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "jelly.h"

int find_maximum_unique(int x, int y, std::vector<int> A, std::vector<int> B) {
	int n = sz(A);
	vector<pii> a(n);
	rep(i,n) a[i] = {A[i],B[i]};
	sort(all(a));
	a.pb({inf1,inf1});
	multiset<int> ms;
	rep(i,n+1) ms.insert(a[i].ss);

	vector<int> dp1(x+5), dp2(x+5);
	dp1[0] = 0;

	int ans = 0;

	rep(i,n+1){
		// i = first ignored
		int mny = dp1[x];
		ms.erase(ms.find(a[i].ss));

		if(mny <= y){
			int cost = i;

			trav(val,ms){
				mny += val;
				if(mny > y) break;
				cost++;
			}

			amax(ans,cost);
		}

		fill(all(dp2),inf1);
		rep(j,x+1){
			// A
			if(j+a[i].ff <= x){
				amin(dp2[j+a[i].ff],dp1[j]);
			}

			// B
			if(dp1[j]+a[i].ss <= y){
				amin(dp2[j],dp1[j]+a[i].ss);
			}
		}

		dp1 = dp2;
		rep1(j,x) amin(dp1[j],dp1[j-1]);
	}

	return ans;

	// int n = a.size();
	// return n;
}

• Subtask 1: 11 / 11
• Subtask 2: 24 / 24
• Subtask 3: 9 / 9
• Subtask 4: 10 / 10
• Subtask 5: 14 / 14
• Subtask 6: 32 / 32