Submission #439997

#TimeUsernameProblemLanguageResultExecution timeMemory
439997elazarkorenColouring a rectangle (eJOI19_colouring)C++17
30 / 100
2091 ms6900 KiB
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#define chkmin(a, b) a = min(a, b)
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> vii;
typedef vector<bool> vb;
typedef vector<vb> vvb;

const ll infinity = 1e18;

int main() {
    int n, m;
    cin >> n >> m;
    vi dig1(m + n - 1), dig2(n + m - 1);
    for (int i = 0; i < m + n - 1; i++) {
        cin >> dig2[i];
    }
    for (int i = 0; i < n + m - 1; i++) {
        cin >> dig1[i];
    }
//    ll ans = 0;
//    for (int i = 0; i < m; i++) {
//        ans += min(dig1[m - i - 1], dig2[i]);
//    }
    ll ans1 = infinity, ans2 = infinity;
    for (int i = 0; i < (1 << ((n + m) / 2)); i++) {
        ll curr = 0;
        vb visit_dig1(n + m - 1);
        for (int j = 0; j < (n + m) / 2; j++) {
            if ((i >> j) & 1) {
                curr += dig1[2 * j];
                visit_dig1[2 * j] = true;
            }
        }
        vb visit_dig2(n + m - 1);
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < m; k++) {
                if (!visit_dig1[j + k] && !((j + k) & 1)) {
                    visit_dig2[j - k + m - 1] = true;
                }
            }
        }
        for (int j = 0; j < n + m - 1; j++) {
            if (visit_dig2[j]) curr += dig2[j];
        }
        chkmin(ans1, curr);
    }
    for (int i = 0; i < (1 << ((n + m - 1) / 2)); i++) {
        ll curr = 0;
        vb visit_dig1(n + m - 1);
        for (int j = 0; j < (n + m - 1) / 2; j++) {
            if ((i >> j) & 1) {
                curr += dig1[2 * j + 1];
                visit_dig1[2 * j + 1] = true;
            }
        }
        vb visit_dig2(n + m - 1);
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < m; k++) {
                if (!visit_dig1[j + k] && ((j + k) & 1)) {
                    visit_dig2[j - k + m - 1] = true;
                }
            }
        }
        for (int j = 0; j < n + m - 1; j++) {
            if (visit_dig2[j]) curr += dig2[j];
        }
        chkmin(ans2, curr);
    }
    cout << ans1 + ans2;
}
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