Submission #59732

#	Time	Username	Problem	Language	Result	Execution time	Memory
59732		model_code	Homecoming (BOI18_homecoming)	C++17	44 / 100	270 ms	110468 KiB

homecoming

// Andrei-Costin Constantinescu
// O(N * K) - Dynamic Programming
#include <bits/stdc++.h>
#include "homecoming.h"

using namespace std;

const int NMAX   = 2000000;
const int VALMAX = 1000000000;

void subtaskAsserts(int N, int K, int a[], int b[]) {
    assert(1LL * N * K <= NMAX);
}

typedef long long int lint;

void asserts(int N, int K, int a[], int b[]) {
    assert(1 <= K && K <= N && N <= NMAX);
    for (int i = 0; i < N; ++ i) {
        assert(0 <= a[i] && a[i] <= VALMAX);
        assert(0 <= b[i] && b[i] <= VALMAX);
    }
}

int A[2 * NMAX + 5], B[2 * NMAX + 5];
lint dpTaken[2 * NMAX + 5];
lint dp[2 * NMAX + 5];
lint sPartB[2 * NMAX];

inline lint getSPartB(int l, int r) {
    lint sum = sPartB[r];
    if (l > 0)
        sum -= sPartB[l - 1];
    return sum;
}

lint solve(int N, int K, int a[], int b[]) {
    asserts(N, K, a, b), subtaskAsserts(N, K, a, b);

    // Copy and double a, b for internal use
    for (int i = 0; i < N; ++ i) {
        A[i] = a[i], A[N + i] = a[i];
        B[i] = b[i], B[N + i] = b[i];
    }

    // Compute the partial sums of B
    sPartB[0] = B[0];
    for (int i = 1; i < 2 * N; ++ i)
        sPartB[i] = sPartB[i - 1] + B[i];

    lint ans = 0;
    const auto f = [&](const int i) {
        dp[i + N] = 0;
        for (int j = i + N - 1; j >= i + 1; -- j) {
            if (j + K - 1 < i + N) {
                dpTaken[j] = dp[j + K];
                if (j + K < i + N)
                    dpTaken[j] = max(dpTaken[j], dpTaken[j + 1] + A[j + 1] - B[j + K]);
            }

            dp[j] = dp[j + 1];
            if (j + K - 1 < i + N)
                dp[j] = max(dp[j], dpTaken[j] + A[j] - getSPartB(j, min(j + K - 1, i + N - 1)));
        }
        ans = max(ans, dp[i + 1]);
    };

    const auto g = [&](const int where_0) {
        dp[where_0 + N] = dpTaken[where_0 + N] = 0;
        for (int j = where_0 + N - 1; j >= where_0; -- j) {
            if (j + K - 1 > where_0 + N - 1) {
                if (j + 1 < where_0 + N)
                    dpTaken[j] = dpTaken[j + 1] + A[j + 1];
                else
                    dpTaken[j] = dpTaken[j + 1];
            }
            else if (j + 1 <= where_0 + N - 1) {
                if (j + K <= where_0 + N - 1)
                    dpTaken[j] = max(dp[j + K], dpTaken[j + 1] + A[j + 1] - B[j + K]);
                else
                    dpTaken[j] = max(dp[j + K], dpTaken[j + 1] + A[j + 1]);
            }
            else
                dpTaken[j] = max(dp[j + K], dpTaken[j + 1]);
            dp[j] = max(dp[j + 1], dpTaken[j] + A[j] - getSPartB(j, min(j + K - 1, where_0 + N - 1)));
        }
        ans = max(ans, dpTaken[where_0] + A[where_0] - getSPartB(where_0, where_0 + K - 1));
    };

    // Fix a textbook we'll NOT take (out of the first K)
    for (int i = 0; i < K; ++ i)
        f(i);

    // Now, assume we take the first K textbooks for sure
    g(0);

    return ans;
}

• Subtask 1: 13 / 13
• Subtask 2: 0 / 18
• Subtask 3: 31 / 31
• Subtask 4: 0 / 38