Submission #1184006

#	Time	Username	Problem	Language	Result	Execution time	Memory
1184006		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;
    
    // Create a DP table where dp[w] represents the maximum value achievable with weight capacity w
    vector<long long> dp(S + 1, 0);
    
    // For each item type
    for (int i = 0; i < N; i++) {
        int value, weight, count;
        cin >> value >> weight >> count;
        
        // Skip if the item is too heavy for the basket
        if (weight > S) continue;
        
        // Use the binary optimization technique
        // Break down count K into powers of 2: 1, 2, 4, 8, ... and a remainder
        // This reduces processing from O(K) to O(log K)
        
        // Create "binary" bundles of items
        vector<pair<long long, int>> bundles;
        for (int c = 1; count > 0; c *= 2) {
            int bundle_size = min(c, count);
            bundles.push_back({(long long)value * bundle_size, weight * bundle_size});
            count -= bundle_size;
        }
        
        // Apply 0/1 knapsack for each bundle
        for (auto& bundle : bundles) {
            long long bundle_value = bundle.first;
            int bundle_weight = bundle.second;
            
            // Skip if the bundle is too heavy
            if (bundle_weight > S) continue;
            
            // Update the DP table from right to left to avoid counting the same bundle multiple times
            for (int w = S; w >= bundle_weight; w--) {
                dp[w] = max(dp[w], dp[w - bundle_weight] + bundle_value);
            }
        }
    }
    
    // Output the maximum value
    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