제출 #497274

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
497274		RandomLB	Knapsack (NOI18_knapsack)	C++17	100 / 100	41 ms	3552 KiB

knapsack

#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef pair<double, double> point;
typedef tuple<int, int, int> tp;  //-------LL
typedef unordered_set<ll> uset;
typedef priority_queue<pll, vector<pll>, greater<pll>> djk;
typedef long long C;
typedef complex<C> P;
#define pb push_back
#define ms(a,b) memset(a,b,sizeof(a))
#define maxE(a) *max_element(a, a+sizeof(a)/sizeof(a[0]))
#define minE(a) *min_element(a, a+sizeof(a)/sizeof(a[0]))
#define F first
#define S second
#define siz(a) (int)a.size()
#define len(a) (int)a.length()
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define bits(x) __builtin_popcount(x)
#define vvll vector<vector<ll>>
#define FOR(a, n) for (int a = 0; a < n; a++)
#define fin(s) {cout << s << "\n"; return;}
#define finm(s) {cout << s << "\n"; return 0;}
#define X real()
#define Y imag()
#define deb(...) logger(#__VA_ARGS__, __VA_ARGS__)
template<typename ...Args>
void logger(string vars, Args&&... values) {
    cout << vars << " = ";
    string delim = "";
    (..., (cout << delim << values, delim = ", "));
    cout << endl;
}
template<class T>
istream& operator >> (istream& is, complex<T>& p) {
  T value;
  is >> value; p.real(value);
  is >> value; p.imag(value);
  return is;
}
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const ll MOD = 1e9+7;
const double pie = 2*acos(0.0);
//------------------------------------------------------------
const int MAX = 2005;
int dp[MAX];
vector<pi> use[MAX];

int main(){ //Key observation: since there only items of up to W 2000, there are bound
    //to be some duplicate weights. We greedily take the largest value for each weight,
    //and we stop considering items of a certain weight after we already have the max
    //possible items(2000/W) of that weight that can be used to fill up the knapsack.
    cin.tie(0)->sync_with_stdio(0);
    int s, n;
    cin >> s >> n;
    bool curr = 1;
    for (int i = 1; i <= n; i++){
        int v, w, k;
        cin >> v >> w >> k;
        use[w].pb({v, k}); //Sort by the values of item
    }
    for (int i = 1; i <= s; i++){
        sort(rall(use[i]));
        int curr = 0; //Keep track of how many items of the weight we have already
        for (int j = 0; j < siz(use[i]); j++){
            curr += use[i][j].S;
            if (curr >= s/i){ //If we already have the max possible items needed, we don't need to consider
                while (siz(use[i]) > j+1) use[i].pop_back(); //the rest of the items with lower value for the same W
                break;
            }
        }
    }
    //Normal knapsack afterwards
    for (int i = 1; i <= s; i++){ //Go through items of each weight
        for (auto[v, k]: use[i]){ //Choose items in powers of 2
            k = min(k, s/i);
            int curr = 0;
            while (k){
                int sz = min((1<<curr), k); curr++;
                for (int j = s; j >= i*sz; j--) dp[j] = max(dp[j], dp[j-i*sz] + sz*v);
                k -= sz;
            }
        }
    }
    cout << dp[s] << "\n";
}

/* stuff you should look for
	* int overflow, array bounds
	* test case inputs don't carry over (DON'T RETURN EARLY)
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/

컴파일 시 표준 에러 (stderr) 메시지

knapsack.cpp: In function 'int main()':
knapsack.cpp:64:10: warning: unused variable 'curr' [-Wunused-variable]
   64 |     bool curr = 1;
      |          ^~~~

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