Submission #903980

#	Time	Username	Problem	Language	Result	Execution time	Memory
903980		nguyen31hoang08minh2003	Holding (COCI20_holding)	C++14	110 / 110	44 ms	10740 KiB

holding

#include <bits/stdc++.h>

/*

    Consider the problem

        Give n integers a[1], ..., a[n]
        and another n integers b[1], ..., b[n]

        Find permutations of (1, ..., n), p and q
            such that |a[p[1]] - b[q[1]]| + ... + |a[p[n]] - b[q[n]]|
                is minimum

        Solution:
            Sort a and b

    Thus, the elements which are originally to the left of the interval [L, R]
        should have final positions to the left to
        final positions of the elements which are originally to the right of the interval [L, R]



*/

template<class A, class B>
bool maximize(A &a, const B& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class A, class B>
bool minimize(A &a, const B& b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

using namespace std;

constexpr int MAX_N = 102, MAX_K = 1E4 + 4, INF = 0X3F3F3F3F, NINF = 0XC0C0C0C0;

int result = INF, N, L, R, K, A[MAX_N], leftCost[MAX_N][MAX_K], rightCost[MAX_N][MAX_K], newCost[MAX_N][MAX_K];

signed main() {
    #ifdef LOCAL
    freopen("input.INP", "r", stdin);
    #endif // LOCAL

    cin.tie(0) -> sync_with_stdio(0);
    cout.tie(0);

    cin >> N >> L >> R >> K;

    for (int i = 1; i <= N; ++i)
        cin >> A[i];

    for (int j = 0, i, k, d; j < L; ++j) {
        for (i = L; i <= R; ++i)
            for (k = 0; k <= K; ++k) {
                newCost[i][k] = newCost[i - 1][k] + A[i];
                if (j >= 1) {
                    d = i - j;
                    minimize(newCost[i][k], leftCost[i][k]);
                    if (k >= d)
                        minimize(newCost[i][k], leftCost[i - 1][k - d] + A[j]);
                }
            }
        for (i = L; i <= R; ++i)
            for (k = 0; k <= K; ++k)
                leftCost[i][k] = newCost[i][k];
    }

    for (int j = N + 1, i, k, d; j > R; --j) {
        for (i = R; i >= L; --i)
            for (k = 0; k <= K; ++k) {
                newCost[i][k] = newCost[i + 1][k] + A[i];
                if (j <= N) {
                    d = j - i;
                    minimize(newCost[i][k], rightCost[i][k]);
                    if (k >= d)
                        minimize(newCost[i][k], rightCost[i + 1][k - d] + A[j]);
                }
            }
        for (i = R; i >= L; --i)
            for (k = 0; k <= K; ++k)
                rightCost[i][k] = newCost[i][k];
    }

    minimize(result, min(leftCost[R][K], rightCost[L][K]));

    for (int i = L, k; i < R; ++i)
        for (k = 0; k <= K; ++k)
            minimize(result, leftCost[i][k] + rightCost[i + 1][K - k]);

    cout << result << '\n';

    return 0;
}

• Subtask 1: 22 / 22
• Subtask 2: 33 / 33
• Subtask 3: 33 / 33
• Subtask 4: 22 / 22