제출 #582502

#제출 시각아이디문제언어결과실행 시간메모리
582502zaneyuKnapsack (NOI18_knapsack)C++14
100 / 100
551 ms10432 KiB
/*input
20 3
5000 15 1
100 1 3
50 1 4
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long, null_type, less<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<ll,ll>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll INF64=4e18;
const int INF=1e6+1;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll sub(ll a,ll b){
    ll x=a-b;
    while(x<0) x+=MOD;
    while(x>MOD) x-=MOD;
    return x;
}
ll mult(ll a,ll b){
    return a*b%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
const ll maxn=100005;
const ll maxlg=__lg(maxn)+2;
vector<ll> dp,new_dp;
struct thing{
    ll v,w,k;
};
vector<thing> w[2005];
bool cmp(thing &a,thing &b){
    return a.v>b.v;
}
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n,s;
    cin>>s>>n;
    vector<pii> v;
    REP(i,n){
        thing b;
        cin>>b.v>>b.w>>b.k;
        w[b.w].pb(b);
    }
    vector<thing> ans;
    REP1(i,2004){
        sort(ALL(w[i]),cmp);
        REP(j,min(sz(w[i]),s/i)){
            ans.pb(w[i][j]);
        }
    }
    REP(i,sz(ans)){
        int k=ans[i].k;
        ll cur=1;
        while(k>0){
            if(k>cur) v.pb({cur*ans[i].v,cur*ans[i].w});
            else v.pb({k*ans[i].v,k*ans[i].w});
            k-=cur;
            cur*=2;
        }
    }  
    dp.resize(2005);
    new_dp.resize(2005);
    n=sz(v);
    dp[0]=0;
    REP1(i,s) dp[i]=-INF;
    REP1(i,n){
        REP(j,s+1){
            new_dp[j]=dp[j];
            if(j>=v[i-1].s){
                MXTO(new_dp[j],dp[j-v[i-1].s]+v[i-1].f);
            }
        }
        dp=new_dp;
    }
    cout<<*max_element(ALL(dp));
}
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