제출 #943385

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
943385		Hadi_Alhamed	Knapsack (NOI18_knapsack)	C++17	100 / 100	82 ms	35176 KiB

knapsack

//to live is to die
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long int ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<ll, ll>> vpl;
#define Clear(a, n)              \
    for (int i = 0; i <= n; i++) \
    {                            \
        a[i] = 0;                \
    }
#define clearMat(a, n, m, d)         \
    for (int i = 0; i <= n; i++)     \
    {                                \
        for (int j = 0; j <= m; j++) \
            a[i][j] = d;             \
    }
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define PB push_back
#define PF push_front
#define MP make_pair
#define F first
#define S second
#define rep(i, n) for (int i = 0; i < n; i++)
#define repe(i, j, n) for (int i = j; i < n; i++)
#define SQ(a) (a) * (a)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define Rrep(i, start, finish) for (int i = start; start >= finish; i--)
#define db(x)  cerr << #x <<" "; _print(x); cerr << endl;

#define forn(i, Start, End, step) for (int i = Start; i <= End; i += step)
#define rforn(i, Start, End, step) for (int i = Start; i >= End; i -= step)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
// ll arr[SIZE];
/*
how to find n % mod ; n < 0?
x = (n+mod)%mod
if(x < 0) x += mod;
*/
void _print(int x)
{
    cerr << x;
}
void _print(ll x)
{
    cerr << x;
}
void _print(string x)
{
    cerr << x;
}
void _print(char x)
{
    cerr << x;
}
void _print(double x)
{
    cerr << x;
}
void _print(ull x)
{
    cerr << x;
}
void _print(vl x)
{
    for(auto e : x)
    {
        cerr << e << " ";
    }
    cerr << "\n";
}
void print(vpi x)
{
    for(auto e : x)
    {
        cerr << e.F << " " << e.S << "\n";
    }
    cerr << "\n";
}
void _print(vi x)
{
    for(auto e : x)
    {
        cerr << e << " ";
    }
    cerr << "\n";
}

void _print(deque<ll>x)
{
    for(auto e : x)
    {
        cerr << e << " ";
    }
    cerr << "\n";
}
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type,  less_equal <T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> bool ckmin(T& a, const T& b)
{
    return b<a?a=b,1:0;
}
template<class T> bool ckmax(T& a, const T& b)
{
    return a<b?a=b,1:0;
}
template<typename T> istream& operator>>(istream& in, vector<T>& a)
{
    for(auto &x : a) in >> x;
    return in;
};
template<typename T> ostream& operator<<(ostream& out, vector<T>& a)
{
    for(auto &x : a) out << x << ' ';
    return out;
};

// priority_queue<data type , the container that would hold the values , greater<pair<int,int>>>
// greater means that we want the smallest value on top
// less means that we want the largest
// x ^ (n) mod m = ( (x mod m)^(n) ) mod m
char to_char(int num)
{
    return (char)(num + '0');
}

ll const MAX = 1e18+1;
ll const oo = 1e18 + 1;
ll const INF = 1e9 + 10;
const ll MOD = 1e9 + 7;
ll const SIZE = 2e5 + 900;
//const int MAX_N = 100'005;
const int LOG = 20;
void solve()
{

    int S , N;
    cin >> S >> N;
    map<int , vpi >by_weight ; // weight : {value , copies}}
    rep(i , N)
    {
        int V , W , K;
        cin >> V >> W >> K;
        K = min(K , S);
        if(W <= S && K >= 1)
        {
            by_weight[W].PB({V , K});
        }
    }
    //dp[i][j] : max value we can achieve using first i weight
    //with weight equal to j
    vector<vector<ll>>dp((int)by_weight.size() + 1, vector<ll>(S + 1 , -1));
    dp[0][0] = 0;
    int at = 1;
    //go weight by weight and see if we take that many copies
    //what is the profit and what is the best answer for the current
    //dp[i][weight]
    ll answer = 0;
    for(auto it = by_weight.begin() ; it != by_weight.end() ; it++)
    {
        vpi items = it->S;
        int w = it->F;
        sort(items.begin(), items.end(), std::greater<pair<int, int>>());

        for(int i = 0; i <= S ;  i++)
        {
            dp[at][i] = dp[at - 1][i];//leave
            answer = max(answer , dp[at][i]);

            int copies = 0;
            ll total_profit = 0;
            int index = 0;
            int copies_of_current_item = 0;
            while((copies + 1) * w <= i && index < (int)items.size())
            {
                total_profit += items[index].F;
                copies_of_current_item++;
                copies++;
                if(i - copies * w >= 0)
                {
                    dp[at][i] = max(dp[at][i] , dp[at - 1][i - copies * w] + total_profit);
                    answer = max(answer , dp[at][i]);
                }
                if(copies_of_current_item == items[index].S)
                {
                    copies_of_current_item = 0;
                    index++;
                }
            }
        }
        at++;
    }
    	cout << *std::max_element(dp.back().begin(), dp.back().end()) << endl;

}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
//    freopen("taming.in" , "r" , stdin);
//    freopen("taming.out", "w" , stdout);

    int T = 1;
//    cin >> T;
    while(T--)
    {
        solve();
    }
    return 0;
}

/* stuff you should look for
 * WRITE STUFF DOWN,  ON PAPER
 * BFS THEN DFS
 * int overflow, array bounds
 * special cases (n=1?)
 * do sm th instead of nothing and stay organized
 * DON'T GET STUCK ON ONE APPROACH
 * (STUCK?)******** Try to simplify the problem(keeping in mind the main problem), ():
 * 1- problem to subProblem
 * 2- from simple to complex: start with a special
 *    problem and then try to update the solution for general case
 *    -(constraints - > solve it with none , one,two ... of them till you reach the given problem
      -(no constraints - > try to give it some)
      -how a special case may be incremented
 * 3-Simplification by Assumptions
 * REVERSE PROBLEM
 * PROBLEM ABSTRACTION
 * SMALL O BSERVATIONS MIGHT HELP ALOT
 * WATCH OUT FOR TIME
 * RETHINK YOUR IDEA,BETTER IDEA, APPROACH?
 * CORRECT IDEA, NEED MORE OBSERVATIONS
 * CORRECT APPROACH, WRONG IDEA
 * WRONG APPROACH
 * THINK CONCRETE THEN SYMBOL,
 * having the solution for the first m state , can we solve it for m + 1 ?
 * in many cases incremental thinking needs data sorting
 */

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