제출 #920663

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
920663		gustavo_d	Knapsack (NOI18_knapsack)	C++17	100 / 100	91 ms	6776 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=0;
    map<int, vector<pair<ll,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<ll,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];
            for (int tmp=0; tmp<K[i]; tmp++) {
                v.push_back({w[i], val[i]});
                tmpn++;
            }
            // 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]});
            //     }
            // }
            // tmpn++;
            // 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: 17 / 17
• Subtask 3: 20 / 20
• Subtask 4: 24 / 24
• Subtask 5: 27 / 27