Submission #976323

#	Time	Username	Problem	Language	Result	Execution time	Memory
976323		stefanopulos	Netrpeljivost (COI23_netrpeljivost)	C++17	100 / 100	513 ms	91148 KiB

Main

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;

#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
 
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const ll inf = 1e15 + 5;
const int mxN = 2050; 

int n;
int mat[mxN][mxN];

ll dp[mxN][mxN];

pii seg[4 * mxN];
int poz[4 * mxN];
void calc(int v, int tl, int tr){
    seg[v] = {tl, tr};
    if(tl == tr){
        poz[tl] = v;
        return;
    }
    int mid = (tl + tr) / 2;
    calc(v * 2, tl, mid); calc(v * 2 + 1, mid + 1, tr);
}

int main(){
    cin.tie(0)->sync_with_stdio(0);

    cin >> n;
    ff(i,0,n - 1)ff(j,0,n - 1)cin >> mat[i][j];

    calc(1,0,n - 1);

    ff(i,0,n)ff(a,0,n)dp[i][a] = inf;
    ff(a,0,n - 1)dp[0][a] = 0;

    ff(i,1,n - 1){

        int X = 0;
        ff(j,0,11){
            if((i >> j) & 1){
                X = j;
                break;
            }
        }

        ff(a,0,n - 1){

            int pos = poz[a];
            pos >>= X;
 
            int L = seg[pos ^ 1].fi;
            int R = seg[pos ^ 1].se;

            ff(b,L,R){
                dp[i][a] = min(dp[i][a], dp[i - 1][b] + mat[a][b]);
            }

        }

    }

    ll rez = inf;
    ff(a,0,n - 1)rez = min(rez, dp[n - 1][a]);
    cout << rez << '\n';

    return 0;
}
/*



// probati bojenje sahovski
*/
 
 
 
 
 

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