Submission #1149789

#	Time	Username	Problem	Language	Result	Execution time	Memory
1149789		monkey133	Netrpeljivost (COI23_netrpeljivost)	C++20	100 / 100	1036 ms	82812 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll INF = 1e15;
ll arr[3005][3005], dp[3005][3005];

int main()
{
	ll n;
	cin >> n;
	for (ll i=0; i<n; ++i) for (ll j=0; j<n; ++j) cin >> arr[i][j];
	for (ll i=0; i<=n; ++i) for (ll j=0; j<=n; ++j) dp[i][j] = INF;
	for (ll i=0; i<n; ++i) dp[0][i] = 0;
	for (ll i=1; i<n; ++i) // no. of len
	{
		ll lsb = i & (-i);
		for (ll j=0; j<n; ++j) // prevv 
		{
			ll st = (j & (~(lsb - 1))); // clear all bit < lsb
			st^=lsb; // switch to different group
			for (ll k=st; k < st + lsb; ++k) // nxt
			{
				dp[i][k] = min(dp[i][k], dp[i-1][j] + arr[j][k]);
			}
		}
	}
	ll ans = *min_element(dp[n-1], dp[n-1] + n);
	cout << ans << endl;

	return 0;
}

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