oj.uz

답안 #171850

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
171850 2019-12-30T13:31:11 Z ne4eHbKa K개의 묶음 (IZhO14_blocks) C++17
0 / 100
 1000 ms 2004 KB 
//{ <defines>
#ifndef _LOCAL
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#endif

#include <bits/stdc++.h>
using namespace std;

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int(x.size())
#define pw(x) (1 << (x))
#define PW(x) (1ll << (x))
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset(x, y, sizeof x)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

const int oo = 2e9;
const ll OO = 4e18;
//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int md = 0x3b800001;
const int MD = 1e9 + 7;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}
//} </defines>
const int N = 1e5 + 5;
const int LG = 18;

int n, k, a[N], lg[N], st[LG][N], c[N];

inline int mx(int l, int r) {
    if(l > r) re 0;
    int f = lg[r - l + 1];
    re max(st[f][l], st[f][r - pw(f) + 1]);
}

void solve() {
    cin >> n >> k;
    fo(n) cin >> a[i];

    copy(a, a + n, st[0]);
    fo(n + 1) lg[i] = i ? lg[i - 1] + (pw(lg[i - 1] + 1) <= i) : 0;
    fo(LG) if(i) for(int j = 0; j + pw(i) - 1 < n; ++j) st[i][j] = max(st[i - 1][j], st[i - 1][j + pw(i - 1)]);

    if(k == 1) re void(cout << mx(0, n - 1) << endl);

    fo(n) {
        int l = 0, r = i + 1;
        while(l < r) {
            int m = (l + r + 1) >> 1;
            mx(i - m, i - 1) >= a[i]
                ? r = m - 1
                : l = m;
        }
        c[i] = l;
    }

    int dp[k - 1][n];
    fo(n) dp[0][i] = mx(0, i);

    fo(k - 1) if(i) {
        fr(j, n) {
            dp[i][j] = oo;
            if(j < i) continue;
            for(int p = j - c[j]; p < j; ++p) if(~p)
                umin(dp[i][j], dp[i - 1][p] + a[j]);
        }
    }

//    fr(j, k - 1) fo(n) cerr << dp[j][i] << (i == n - 1 ? '\n' : ' ');

    int ans = oo;

    fo(n - 1) umin(ans, dp[k - 2][i] + mx(i + 1, n - 1));

    cout << ans << endl;
}

int main() {
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    fo(tests) {
		cerr << "case #" << i+1 << endl;
		solve();
	}
#else
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
	solve();
#endif
	return 0;
}

Subtask #1
0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 376 KB Output is correct
3 Correct 2 ms 376 KB Output is correct
4 Correct 2 ms 376 KB Output is correct
5 Correct 2 ms 296 KB Output is correct
6 Correct 2 ms 376 KB Output is correct
7 Correct 2 ms 376 KB Output is correct
8 Correct 2 ms 376 KB Output is correct
9 Incorrect 2 ms 376 KB Output isn't correct
10 Halted 0 ms 0 KB -

Subtask #2
0 / 18.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 380 KB Output is correct
2 Correct 2 ms 376 KB Output is correct
3 Correct 2 ms 376 KB Output is correct
4 Correct 2 ms 376 KB Output is correct
5 Correct 2 ms 376 KB Output is correct
6 Correct 2 ms 376 KB Output is correct
7 Correct 2 ms 376 KB Output is correct
8 Correct 2 ms 376 KB Output is correct
9 Incorrect 2 ms 376 KB Output isn't correct
10 Halted 0 ms 0 KB -

Subtask #3
0 / 21.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 2 ms 376 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #4
0 / 47.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 1070 ms 2004 KB Time limit exceeded
2 Halted 0 ms 0 KB -