Submission #1274041

#	Time	Username	Problem	Language	Result	Execution time	Memory
1274041		quanduongxuan12	Knapsack (NOI18_knapsack)	C++20	100 / 100	41 ms	2980 KiB

knapsack

#include <bits/stdc++.h>
using namespace std;
#define name "knapsack"
#define MAXN 100005
#define pb push_back
#define pf push_front
#define ll long long
#define ii pair<int, int>
#define fs first
#define sc second
#define ill pair<int, ll>
#define lli pair<ll, int>
#define llll pair<ll, ll>
#define all(v) v.begin(),v.end()
#define uni(v) v.erase(unique(all(v)),v.end())
#define bit(n,i) (((n)>>(i))&1)
#define FOR(i,a,b) for (int i=(a),_b=(b); i<=_b; i++)
#define FORD(i,a,b) for (int i=(a),_b=(b); i>=_b; i--)
#define MASK(i) (1LL<<(i))
const int INF=1e9;
const int MOD=1e9+7;
void add(int &u, int v){
    u+=v;
    if (u>=MOD) u-=MOD;
}
void sub(int &u, int v){
    u-=v;
    if (u<0) u+=MOD;
}
void minimize(int &u, int v){
    u=min(u,v);
}
void maximize(ll &u, ll v){
    u=max(u,v);
}
long long Rand(long long l, long long r){
    ll tmp=0;
    FOR(i,1,4) tmp=((tmp<<15)^(((1<<15)-1)&rand()));
    return l+tmp%(r-l+1);
}
int S,n;
int V[MAXN],W[MAXN],K[MAXN];
vector<ii> store[2003];
vector<ll> f[2003];
ll dp[2][2003];
int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin>>S>>n;
    FOR(i,1,n) cin>>V[i]>>W[i]>>K[i];
    FOR(i,1,n){
        store[W[i]].pb({V[i],K[i]});
    }
    FOR(i,1,S){
        sort(all(store[i]));
        int LIM=S/i;
        int j=1;
        f[i].pb(0);
        while (store[i].size()){
            ii tmp=store[i].back();
            store[i].pop_back();
            while (tmp.sc>0&&j<=LIM){
                f[i].pb(f[i].back()+1LL*tmp.fs);
                ++j;
                --tmp.sc;
            }
            if (j>LIM) break;
        }
    }
    /*
    FOR(i,1,S){
        for (int j:f[i]) cout<<j<<" ";
        cout<<"\n";
    }
    */
    FOR(i,1,S){
        FOR(j,1,S){
            maximize(dp[i&1][j],dp[i&1][j-1]);
            FOR(x,0,min((int)j/i,(int)f[i].size()-1)){
                maximize(dp[i&1][j],dp[(i&1)^1][j-x*i]+f[i][x]);
            }
        }
    }
    cout<<dp[S&1][S];
    return 0;
}

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