제출 #391662

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
391662		saarang123	K개의 묶음 (IZhO14_blocks)	C++17	0 / 100	1 ms	336 KiB

blocks

#include <bits/stdc++.h>
using namespace std;
const int mxn = 100 * 1000 + 3, mxk = 103;
int dp[mxk][mxn], a[mxn];
int n, k;

template<class T> struct SparseTable { // comb(ID,b) = b
    T ID = T();
    function<T(const T&,const T&)> comb;
    int n, k; vector<vector<T>> seg;
    void init(int _n, T base, function<T(const T&,const T&)> join) {
        n = _n; k = lg(n); comb = join; ID = base; seg.assign(_n + 1, vector<T> (k + 1, ID)); }
    int lg(int x) { return 31 - __builtin_clz(x); }   //CHECK THIS
    void build() {
        for(int i = 1; i <= n; i++) seg[i][0] = a[i];
        for(int j = 1; j <= k; j++)
            for(int i = 1; i + (1 << j) - 1 <= n; i++)
                seg[i][j] = comb(seg[i][j - 1], seg[i + (1 << (j - 1))][j - 1]);
    }
    T query(int L, int R) {
        int j = lg(R - L + 1);
        return comb(seg[L][j], seg[R - (1 << j) + 1][j]);
    }
};

SparseTable<int> st;

void f(int i, int l, int r, int optl, int optr) {
	if(l > r)
		return;
	int mid = (l + r) >> 1, opt = -1;
	dp[i][mid] = 1e9;
	for(int m = optl; m < min(optr, mid); m++) {
		int cost = st.query(m + 1, mid);
		if(dp[i][mid] > dp[i - 1][m] + cost) {
			dp[i][mid] = dp[i - 1][m] + cost;
			opt = m;
		}
	}
	//cout << i << ' ' << mid << ' ' << dp[i][mid] << ' ' << opt << '\n';
	f(i, l, mid - 1, optl, opt + 1);
	f(i, mid + 1, r, opt, optr);
}

signed main() {
    std::ios::sync_with_stdio(0);
    std::cout.tie(0);
    std::cin.tie(0);
    #ifdef saarang
    freopen("/home/saarang/Documents/cp/input.txt", "r", stdin);
    freopen("/home/saarang/Documents/cp/output.txt", "w", stdout);
    #endif
    cin >> n >> k;
    for(int i = 0; i <= n; i++)
    	for(int j = 0; j <= k; j++)
    		dp[i][j] = 1e9;
    dp[0][0] = 0;
    for(int i = 1; i <= n; i++) 
    	cin >> a[i];
    st.init(n, 0, [&] (int x, int y) {
    	return max(x, y);
    });
    st.build();
    for(int i = 1; i <= k; i++)
    	f(i, 1, n, 0, n);
    // for(int i = 1; i <= k; i++) {
    // 	for(int j = 1; j <= n; j++)
    // 		cout << dp[i][j] << ' ';
    // 	cout << '\n';
    // }
    cout << dp[k][n] << '\n';
    return 0;
}

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