| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 943374 | Hadi_Alhamed | Knapsack (NOI18_knapsack) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//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)
{
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<int>>dp(N + 2 , vector<int>(S + 2 , -1));
dp[0][0] = 1;
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& [w , items] : by_weight)
{
sort(rall(items));
for(int i = 0; i <= S ; i++)
{
dp[at][i] = dp[at - 1][i];//leave
int copies = 0;
int total_weight = 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;
total_weight += w;
copies_of_current_item++;
copies++;
if(i - total_weight >= 0)
{
dp[at][i] = max(dp[at][i] , dp[at - 1][i - total_weight] + total_profit);
answer = max(answer , dp[at][i]);
}
if(copies_of_current_item == items[index].S)
{
copies_of_current_item = 0;
index++;
}
}
}
at++;
}
}
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
*/