dpB - Frog 2

using namespace std;
const int NMAX = 1e5 + 11;
int n, k, h[NMAX], mem[NMAX];
bool vis[NMAX];
inline int dh(int i, int j) { return (int)abs(h[i] - h[j]); }
int dp(int i)
{
    int &ans = mem[i];
    if (i == n - 1)
        return 0;
    if (vis[i])
        return ans;
    vis[i] = true;
    ans = INT_MAX;
    for (int j = i + 1; j <= i + k; ++j)
        if (j < n)
            ans = min(ans, dp(j) + dh(i, j));
    return ans;
}
int main(void)
{
    cin >> n >> k;
    for (int i = 0; i < n; ++i)
        cin >> h[i];
    cout << dp(0) << endl;
    return 0;
}