Submission #954488

#	Time	Username	Problem	Language	Result	Execution time	Memory
954488		GrandTiger1729	Netrpeljivost (COI23_netrpeljivost)	C++17	59 / 100	1522 ms	82732 KiB

Main

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;

const long long INF = 1e18;
int main()
{
    cin.tie(0)->sync_with_stdio(0);
    int n;
    cin >> n;
    vector<vector<int>> a(n, vector<int>(n));
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            cin >> a[i][j];
        }
    }
    if (n == 1)
    {
        cout << 0 << '\n';
        return 0;
    }
    vector<vector<long long>> dp(n, vector<long long>(n, INF)),
                             dp2(n, vector<long long>(n, INF));
    function<void(int, int)> solve = [&](int l, int r)
    {
        if (l + 2 == r)
        {
            dp[l][r - 1] = dp[r - 1][l] = a[l][r - 1];
            return;
        }
        int mid = (l + r) / 2;
        solve(l, mid);
        solve(mid, r);
        for (int i = l; i < (l + mid) / 2; i++)
        {
            for (int j = (l + mid) / 2; j < mid; j++)
            {
                for (int k = mid; k < r; k++)
                {
                    dp2[i][k] = min(dp2[i][k], dp[i][j] + a[j][k]);
                    dp2[j][k] = min(dp2[j][k], dp[j][i] + a[i][k]);
                }
            }
        }
        for (int i = l; i < mid; i++)
        {
            for (int j = mid; j < (mid + r) / 2; j++)
            {
                for (int k = (mid + r) / 2; k < r; k++)
                {
                    dp[i][k] = min(dp[i][k], dp2[i][j] + dp[j][k]);
                    dp[i][j] = min(dp[i][j], dp2[i][k] + dp[k][j]);
                }
            }
        }
        for (int i = l; i < mid; i++)
        {
            for (int j = mid; j < r; j++)
            {
                dp[j][i] = min(dp[j][i], dp[i][j]);
            }
        }
    };
    solve(0, n);
    long long ans = INF;
    for (int i = 0; i < n / 2; i++)
    {
        for (int j = n / 2; j < n; j++)
        {
            ans = min(ans, dp[i][j]);
            ans = min(ans, dp[j][i]);
        }
    }
    cout << ans << '\n';
    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 17 / 17
• Subtask 3: 32 / 32
• Subtask 4: 0 / 41