Submission #543487

#	Time	Username	Problem	Language	Result	Execution time	Memory
543487		SavicS	K blocks (IZhO14_blocks)	C++17	0 / 100	1 ms	340 KiB

blocks

#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 ld;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> 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 pb push_back
#define fi first
#define se second
#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 int inf = 1e9 + 5;
const int mxN = 200005; 

int n, k;
int niz[mxN];

int sp[mxN][20];
int rmq(int l, int r){
    int x = 31 - __builtin_clz(r - l + 1);
    return max(sp[l][x], sp[r - (1 << x) + 1][x]);
}

int dp[mxN][105];
void divi(int s, int l, int r, int tl, int tr) {
    if(l > r)return;
    int mid = (l + r) / 2;

    int opt = 0;
    ff(i,tl,min(mid,tr)) {
        if(dp[i - 1][s - 1] + rmq(i, mid) < dp[mid][s]) {
            dp[mid][s] = dp[i - 1][s - 1] + rmq(i, mid);
            opt = i;
        }
    }
 
    divi(s, l, mid - 1, tl, opt);
    divi(s, mid + 1, r, opt, tr);
 
}

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

    cin >> n >> k;
    ff(i,1,n)cin >> niz[i];

    ff(i,1,n)sp[i][0] = niz[i];
    for(int j = 1; j < 20; j++){
        for(int i = 1; i + (1 << j) - 1 <= n; i++){
            sp[i][j] = max(sp[i][j - 1], sp[i + (1 << (j - 1))][j - 1]);
        }
    }

    ff(i,0,n)ff(j,0,k)dp[i][j] = inf;

    dp[0][0] = 0;
    ff(i,1,k)divi(i,1,n,1,n);

    cout << dp[n][k] << '\n';

    return 0;
}
/*
 
5 1
1 2 3 4 5 

5 2
1 2 3 4 5
 
// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 0 / 14
• Subtask 2: 0 / 18
• Subtask 3: 0 / 21
• Subtask 4: 0 / 47