DMOJ · cco10p4

Computer Purchase Return

This C++ solution uses grouped knapsack for DMOJ cco10p4 Computer Purchase Return. Read the reasoning, inspect the code, or try your own test case below.

cco10p4Dynamic programmingGrouped knapsackC++37 lines
Solution050of 248
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Approach

Grouped knapsack

Computer Purchase Return groups components by type and asks for maximum value under budget using one of every type; the code uses grouped knapsack.

Dynamic programming

Problem and code

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Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

P_4_Computer_Purchase_Return.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1e18; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int t, n; cin >> t >> n;
        vector<vector<pair<int, int>>> group(t + 1);
        for (int i = 0; i < n; i++) {
            int c, v, tt; cin >> c >> v >> tt;
            group[tt].emplace_back(c, v);
        }
        int b; cin >> b;
        for (int i = 1; i <= t; i++) {
            if (group[i].empty()) {
                cout << -1 << '\n';
                return 0;
            }
        }
        vector<int> old(b + 1, -inf);
        old[0] = 0;
        for (int i = 1; i <= t; i++) {
            vector<int> cur(b + 1, -inf);
            for (auto[c, v] : group[i]) {
                for (int j = c; j <= b; j++) {
                    if (old[j - c] == -inf) continue;
                    cur[j] = max(cur[j], old[j - c] + v);
                }
            }
            swap(old, cur);
        }
        int ans = *max_element(old.begin(), old.end());
        cout << (ans < 0 ? -1 : ans) << '\n';
        return 0;
    }
        

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