제출 #1317909

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1317909		pancankes	Knapsack (NOI18_knapsack)	C++17	100 / 100	68 ms	33268 KiB

knapsack

#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <array>
#include <cmath>
using namespace std;
using ll = long long;

const int MAXN = 1e5+1;
const int MAXW = 2e3+1;
const int NINF = -1e9;

void setIO(string name="") { 
	ios_base::sync_with_stdio(0); cin.tie(0); 
}

// naive approach: put k copies of everythign into one vector, run standard 0/1 knapsack
// fails on K <= 1e9, 
// W <= 2000. N*W means that naive 0/1 fails anyways. 

void solve()
{
    int x,n;
    cin >> x >> n;
    map<int,vector<pair<int,int>>> by_w;
    for (int i=1; i<=n; i++) {
        int v,w,k;
        cin >> v >> w >> k;
        by_w[w].push_back({v,k});
    }
    for (int i=0; i<=n; i++) {

    }
    vector<vector<ll>> dp(by_w.size() + 1,vector<ll>(x + 1, NINF)); // initiliase all values of dp vector as INT32_min
    dp[0][0]=0;
    int at=1;
    // grouped the items by weight, sort by decreasing value
    // only need to consider the first x/w items
    for (auto &[w,items] : by_w) {
        sort(items.rbegin(),items.rend());
        for (int j=0; j<=x; j++) {
            dp[at][j]=dp[at-1][j];

            // implementation: go through as many items at weight w
            // until we have no space: (copies + 1) * w <= j OR;
            // until we have no more copies at weight w

            ll copies=0, type=0, cur=0, totalv=0;
            while ((copies+1) * w <= j && type < items.size()){
                copies++;
                totalv+=items[type].first;
                if (dp[at-1][j-copies*w] != NINF) { // here we need a check to see if it's even possible to make dp[i][j];
                    dp[at][j]=max(dp[at][j],dp[at-1][j-copies*w]+totalv);
                } 

                // need to implement the logic of moving on to the next item
                cur++;
                if (cur==items[type].second) {
                    type++;
                    cur=0;
                }
            } 
        }
        at++;
    }
    cout << *max_element(dp.back().begin(),dp.back().end()) << endl;
}

int main() 
{
    setIO();
    solve();
}

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