Submission #517015

#	Time	Username	Problem	Language	Result	Execution time	Memory
517015		DiTenix	Knapsack (NOI18_knapsack)	C++17	0 / 100	1 ms	332 KiB

knapsack

#define eb emplace_back
#define pb push_back
#define pii pair<int,int>
#define sz(x) int((x).size())
#define ALL(x) (x).begin(),(x).end()
#define ln cout << '\n'
#define REP(i, a) for (int i = 0; i < int(a); i++)
#define FOR(i, a) for (int i = 1; i <= int(a); i++)
#define MEM(a,b) memset((a),(b),sizeof(a))
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
using namespace std;
#include <bits/stdc++.h>
using ll = long long;
using vi = vector<int>;
template<typename T> void rd(vector<T> &a) {for (auto &ele : a) cin >> ele;}
template<typename T> void writeln(vector<T> &a) {for (auto &ele : a) cout << ele << ' '; cout << "\n";}
#if __cplusplus > 201402L
template<typename... Args> void rd(Args&... args) {((cin >> args), ...);}
template<typename... Args> void write(Args... args) {((cout << args << " "), ...);}
template<typename... Args> void writeln(Args... args) {((cout << args << " "), ...); cout << "\n";}
#endif
const int maxn = 1e6 + 5;

ll v[maxn], w[maxn], k[maxn];
ll dp[2005][2005];
struct item {
    int v, k;
};
vector<item> b[2005];
signed main()
{

    cin.tie(0); cout.tie(0);
    ios::sync_with_stdio(false);
//    int t;cin >> t;while(t--) solve();
    int n, S;
    cin >> S >> n;
    FOR(i, n) {
        int v, w, k;
        rd(v, w, k);
        b[w].pb({v, k});
    }
    int at = 0;
    for (int i = 1; i <= S; ++i) {
        if (b[i].empty()) continue;
        at++;
        sort(ALL(b[i]), [](item & a, item & b) {
            return a.v > b.v;
        });
        vector<pair<ll, ll>> can_take; // (weight, value)
        ll weight = 0, value = 0;
        for (auto [v, k] : b[i]) {
            for (int c = 0; c < k; ++c) {
                weight += i;
                value += v;
                if (weight >= S) break;
                can_take.eb(weight, value);
            }
            if (weight >= S) break;
        }
        if (!can_take.empty()) {
            for (int j = S; j >= 1; --j) {
                dp[at][j] = dp[at - 1][j];
                for (auto [w, v] : can_take) {
                    if (w > j) {
                        break;
                    }
                    dp[at][j] = max(dp[at][j], dp[at - 1][j - w] + v);
                }
            }
        }
    }
    ll res = 0;
    for (int i = 0; i <= S; ++i) {
        res = max(res, dp[at][i]);
    }
    writeln(res);
    return 0;
}

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