Submission #1208134

#	Time	Username	Problem	Language	Result	Execution time	Memory
1208134		gatapdev	K blocks (IZhO14_blocks)	C++20	0 / 100	1 ms	468 KiB

blocks

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int INF = (int)1e18;

int n, K;
vector<int> a;
vector<vector<int>> dp;
vector<vector<int>> st;
vector<int> logTable;

void buildSparseTable() {
    int maxLog = floor(log2(n)) + 1;
    st.assign(maxLog, vector<int>(n + 1));
    logTable.resize(n + 1);

    logTable[1] = 0;
    for (int i = 2; i <= n; ++i)
        logTable[i] = logTable[i/2] + 1;

    for (int i = 1; i <= n; ++i)
        st[0][i] = a[i];

    for (int k = 1; k < maxLog; ++k) {
        for (int i = 1; i + (1 << k) - 1 <= n; ++i) {
            st[k][i] = max(st[k-1][i], st[k-1][i + (1 << (k-1))]);
        }
    }
}

inline int getMax(int l, int r) {
    int len = r - l + 1;
    int k = logTable[len];
    return max(st[k][l], st[k][r - (1<<k) + 1]);
}

// Compute dp[k][i] for i in [l..r], knowing the optimal split j for dp[k][mid] ∈ [optL..optR]
void compute(int k, int l, int r, int optL, int optR) {
    if (l > r) return;
    int mid = (l + r) >> 1;

    // tìm j tốt nhất trong [optL .. min(mid-1, optR)]
    int bestJ = optL;
    int64_t bestCost = INF;
    int upper = min((int)mid - 1, optR);
    for (int j = optL; j <= upper; ++j) {
        int64_t cost = dp[k-1][j] + getMax(j+1, mid);
        if (cost < bestCost) {
            bestCost = cost;
            bestJ = j;
        }
    }
    dp[k][mid] = bestCost;

    // chia để trị
    compute(k, l, mid-1, optL, bestJ);
    compute(k, mid+1, r, bestJ, optR);
}

int32_t main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n >> K;
    a.resize(n+1);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }

    // 1) build sparse table để query max O(1)
    buildSparseTable();

    // 2) khởi dp
    dp.assign(K+1, vector<int>(n+1, INF));
    // với k=1, chỉ có 1 block, dp[1][i] = max(a[1..i])
    for (int i = 1; i <= n; ++i) {
        dp[1][i] = getMax(1, i);
    }

    // 3) tính cho k = 2..K bằng divide & conquer optimization
    for (int k = 2; k <= K; ++k) {
        // đảm bảo chúng ta chỉ xét j >= k-1
        compute(k, /*l=*/1, /*r=*/n,
                /*optL=*/k-1, /*optR=*/n-1);
    }

    cout << dp[K][n] << "\n";
    return 0;
}

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