Submission #920661

#	Time	Username	Problem	Language	Result	Execution time	Memory
920661		gustavo_d	Knapsack (NOI18_knapsack)	C++17	12 / 100	0 ms	348 KiB

knapsack

// https://oj.uz/problem/view/NOI18_knapsack > p83 e p199
#include <bits/stdc++.h>
using namespace std;

#pragma GCC optimize("03")

typedef long long ll;

const int MAXS = 2000;
const int MAXLOG = 12;

ll dp[2][MAXS+1];

int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
 
    int S, n; cin >> S >> n;
    vector<ll> val(n);
    vector<int> w(n);
    vector<int> K(n);
    vector<pair<int, ll>> v; // peso, valor
    v.reserve(n*MAXLOG);
    int tmpn=n;
    map<int, vector<pair<int,int>>> by_w;
    for (int i=0; i<n; i++) {
        cin >> val[i] >> w[i] >> K[i];
        K[i] = min(K[i], (S/w[i]));

        by_w[w[i]].push_back({val[i], i});    
    }
    for (pair<int,vector<pair<int,int>>> eq_w : by_w) {
        int limit = S / eq_w.first;
        sort(eq_w.second.begin(), eq_w.second.end());
        for (pair<int,int> idx : eq_w.second) {
            int i = idx.second;
            K[i] = min(K[i], limit);
            limit -= K[i];
            if (K[i] == 0) continue;
            bool fi = true;
            int other = K[i];
            for (int b=MAXLOG; b>=0; b--) {
                if (fi and ((1 << b) & K[i]) != 0) {
                    fi = false;
                    other -= ((1 << b) - 1);
                } else if (!fi) {
                    tmpn++;
                    v.push_back({(1 << b) * w[i], (1LL << b) * val[i]});
                }
            }
            v.push_back({other * w[i], other * val[i]});
        }
    }

    for (int i=1; i<=tmpn; i++) {
        for (int P=1; P<v[i-1].first; P++) {
            dp[i&1][P] = max(dp[(i&1)^1][P], dp[i&1][P-1]);
        }
        for (int P=v[i-1].first; P<=S; P++) {
            dp[i&1][P] = max({
                dp[(i&1)^1][P-v[i-1].first] + v[i-1].second,
                dp[(i&1)^1][P],
                dp[i&1][P-1]
            });
        }
    }
 
    cout << dp[tmpn&1][S] << endl;
 
    return 0;
}

• Subtask 1: 12 / 12
• Subtask 2: 0 / 17
• Subtask 3: 0 / 20
• Subtask 4: 0 / 24
• Subtask 5: 0 / 27