Submission #1201398

#	Time	Username	Problem	Language	Result	Execution time	Memory
1201398		tdkhai	Knapsack (NOI18_knapsack)	C++20	29 / 100	1 ms	328 KiB

knapsack

/*
        _.-- ,.--.
      .'   .'     /
       @       |'..--------._
     /      \._/              '.
    /  .-.-                     \
   (  /    \                     \
   \\      '.                  | #
    \\       \   -.           /
     :\       |    )._____.'   \
      "       |   /  \  |  \    )
              |   |./'  :__ \.-'
              '--'
*/
#include<bits/stdc++.h>
#define ll long long
#define llll pair<int,int>
#define ii pair<int,int>
#define fi first
#define se second
#define FOR(i,l,r) for(int i=l;i<=r;i++)
#define FOD(i,r,l) for(int i=r;i>=l;i--)
#define ull unsigned long long
#define iii pair<int,ii>
#define iv pair<pii,ii>
#define db double
#define ld long double
#define pb push_back


using namespace std;

const int dx[]  = {1, -1, 0, 0};
const int dy[]  = {0, 0, -1, 1};
const int dxx[] = {1, 1, -1, -1, 2, 2, -2, -2};
const int dyy[] = {2, -2, 2, -2, 1, -1, 1, -1};
const ll INF=1e18;


ll s,n,ans;
ll dp[2001];
vector<pair<ll,ll>> vec;
void solve()
{
    cin >> s >> n;
    for(int i=1;i<=n;i++)
    {
        ll v,w,k;
        cin >> v >> w >> k;
        int bp=1;
        k=min(k,s/w);
        while(k>=bp)
        {
            vec.pb({w*bp,v*bp});
            k-=bp;
            bp*=2;
        }
        if(k)
        {
            vec.pb({w*k,v*k});
        }
    }
    sort(vec.begin(),vec.end());
    vec.erase(unique(vec.begin(),vec.end()),vec.end());
    n=vec.size();
    for(int i=1;i<=n;i++)
    {
//        cout << vec[i-1].fi << ' ' << vec[i-1].se << '\n';
        for(int j=s;j>=vec[i-1].fi;j--)
        {
            dp[j]=max(dp[j],dp[j-vec[i-1].fi]+vec[i-1].se);
        }
    }
    for(int i=0;i<=s;i++)
    {
        ans=max(ans,dp[i]);
    }
    cout << ans;
}
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//    freopen("tdk.inp","r",stdin);
//    freopen("tdk.out","w",stdout);
    int t=1;
    //cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

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