LeetCode 1883. Minimum Skips to Arrive at Meeting On Time

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const int N = 1001;
const double eps = 1e-8, INF = INT_MAX;
double f[N][N];
class Solution {
public:
    int minSkips(vector<int>& dist, int speed, int hoursBefore) {
        int n = dist.size();
        for (int i = 1; i <= n; ++i) {
            double t = (double) dist[i - 1] / speed;
            for (int j = 0; j <= i; ++j) {
                f[i][j] = INF;
                if (j <= i - 1) f[i][j] = ceil(f[i - 1][j] + t - eps);
                if (j) f[i][j] = min(f[i][j], f[i - 1][j - 1] + t);
            }
        }
        for (int i = 0; i <= n; ++i)
            if (f[n][i] < hoursBefore + eps)
                return i;
        return -1;
    }
};