Submission #1184005

#	Time	Username	Problem	Language	Result	Execution time	Memory
1184005		ayushtiwari110	Knapsack (NOI18_knapsack)	C++20	37 / 100	1 ms	328 KiB

knapsack

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    // Read the maximum capacity S and number of item types N
    int S, N;
    cin >> S >> N;
    
    // Input arrays
    vector<int> values(N);   // Vi: value of item i
    vector<int> weights(N);  // Wi: weight of item i
    vector<int> counts(N);   // Ki: number of copies of item i available
    
    // Read item information
    for (int i = 0; i < N; i++) {
        cin >> values[i] >> weights[i] >> counts[i];
    }
    
    // Create a DP table where dp[w] represents the maximum value achievable with a weight capacity of w
    vector<int> dp(S + 1, 0);
    
    // For each item type
    for (int i = 0; i < N; i++) {
        // Process each item using the binary optimization technique
        // We'll convert K copies into powers of 2 + a remaining item
        int value = values[i];
        int weight = weights[i];
        int count = counts[i];
        
        // Create "binary" copies of the item to handle efficiently
        vector<pair<int, int>> binary_items;
        
        // Express count in binary representation (1, 2, 4, 8, etc.)
        for (int k = 1; count > 0; k *= 2) {
            int take = min(k, count);
            if (take > 0) {
                binary_items.push_back({take * value, take * weight});
            }
            count -= take;
        }
        
        // Apply 0/1 knapsack for each binary item
        for (auto& item : binary_items) {
            int item_value = item.first;
            int item_weight = item.second;
            
            // Update the DP table from right to left to avoid counting the same item multiple times
            for (int w = S; w >= item_weight; w--) {
                dp[w] = max(dp[w], dp[w - item_weight] + item_value);
            }
        }
    }
    
    // The answer is in dp[S]
    cout << dp[S] << endl;
    
    return 0;
}

• Subtask 1: 0 / 12
• Subtask 2: 17 / 17
• Subtask 3: 20 / 20
• Subtask 4: 0 / 24
• Subtask 5: 0 / 27