제출 #1211899

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1211899		omarpaladines95	나일강 (IOI24_nile)	C++20	17 / 100	2095 ms	4728 KiB

nile

#include <bits/stdc++.h>

// This obtains 47 points.

using namespace std;

vector<long long> calculate_costs(
    vector<int> W, vector<int> A,
    vector<int> B, vector<int> E) {

    int N = W.size();
    int Q = E.size();
    
    vector<tuple<int, int, int>> artifacts(N);
    for (int i = 0; i < N; ++i) {
        artifacts[i] = make_tuple(W[i], A[i], B[i]);
    }

    // Sort artifacts by weight ascending first
    sort(artifacts.begin(), artifacts.end());

    /*
    for (int i = 0; i < artifacts.size(); ++i)
        {
            cout << get<0>(artifacts[i]) << " " <<  get<1>(artifacts[i]) << " " <<  get<2>(artifacts[i]) << endl;
        }
    */

    // Results for each D in E
    vector<long long> results(Q);

    for (int q = 0; q < Q; ++q) {
        int D = E[q];

        // DP[i] stores the minimum cost for sending items from 0 to i.
        vector<long long> dp(N, ULLONG_MAX);
        dp[0] = get<1>(artifacts[0]); // Base case: Send 0 on its own.

        for (int i = 1; i < N; ++i) {
            // Case 1: send artifact i alone
            dp[i] = get<1>(artifacts[i]) + dp[i-1];

            // Case 2: try to pair i with with everything below it that's possible
            // By sending everything in between on its own. Them being paired will be considered on their turn of the DP.
            long long a_Cost_In_Between = 0;
            for (int j = i-1; j >= 0; --j) {
                int i_weight = get<0>(artifacts[i]);
                int j_weight = get<0>(artifacts[j]);
                if (i_weight - j_weight > D) break;
                
                // Accumulate the A cost of the artifacts in the range j, [j+1, ... , i-1] , i
                int aCost = 0;
                if (j+1 < i)
                {
                    aCost = get<1>(artifacts[j+1]);
                }
                a_Cost_In_Between += aCost;

                long long costFromPairing = get<2>(artifacts[i]) + get<2>(artifacts[j]) + a_Cost_In_Between;
                long long leftOverCost = j-1 >= 0 ? dp[j-1] : 0;
                dp[i] = min(dp[i], costFromPairing + leftOverCost);
            }
            //cout << dp[i] << endl;
        }

        results[q] = dp[N-1];
        //std::cout << results[q] << endl;
    }

    return results;
}

• Subtask 0: 0 / 0
• Subtask 1: 6 / 6
• Subtask 2: 0 / 13
• Subtask 3: 0 / 17
• Subtask 4: 11 / 11
• Subtask 5: 0 / 20
• Subtask 6: 0 / 15
• Subtask 7: 0 / 18