Submission #14804

#TimeUsernameProblemLanguageResultExecution timeMemory
14804erdemkirazK blocks (IZhO14_blocks)C++98
100 / 100
195 ms43904 KiB
#include <algorithm>
#include <iostream>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

#define type(x) __typeof((x).begin())
#define foreach(i, x) for(type(x) i = (x).begin(); i != (x).end(); i++)

typedef long long ll;
typedef pair < int, int > ii;

const int inf = 1e9 + 333;
const ll linf = 1e18 + 333;

const int N = 1e5 + 5;
const int K = 100 + 5;

int n, k, a[N], dp[N][K];
ii p[N];

int main () {
    
    ios :: sync_with_stdio(0);
    
    cin >> n >> k;
    
    int mx = 0;
    
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
        mx = max(mx, a[i]);
        dp[i][1] = mx;
    }
    
    for(int j = 2; j <= k; j++) {
        int sz = 0;
        for(int i = 1; i <= n; i++) {
            int ans = j <= i ? dp[i - 1][j - 1] : inf;
            while(sz and a[i] >= p[sz - 1].first)
                ans = min(ans, p[--sz].second);
            if(!sz or a[i] + ans < p[sz - 1].first + p[sz - 1].second)
                p[sz++] = ii(a[i], ans);
            dp[i][j] = p[sz - 1].first + p[sz - 1].second;
        }
    }
    
    cout << dp[n][k] << endl;
    
    return 0;
    
}
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