On a projective surface $S$, a $\mathbb{Z}$-divisor $D$ admits a unique Zariski decomposition , $$ D = P + N \,, $$ where $P$ and $N$ are $\mathbb{Q}$-divisors called the 'positive' and 'negative' parts. The important aspect for my question is that $P$ is nef and contains the information about the zeroth cohomology of the line bundle $\mathcal{O}(D)$, in the sense that $$ H^0(S,\mathcal{O}(D)) \cong H^0(S,\mathcal{O}(\lfloor P \rfloor)) \,. $$ Here $\lfloor P \rfloor$ is defined by taking the floor of every coefficient in the divisor expansion of $P$.
If the higher cohomologies of $\mathcal{O}(\lfloor P \rfloor)$ vanish, i.e. $h^q(S,\mathcal{O}(\lfloor P \rfloor)) = 0$ for $q>0$, then $$ h^0(S,\mathcal{O}(D)) = h^0(S,\mathcal{O}(\lfloor P \rfloor)) = \chi(\mathcal{O}(\lfloor P \rfloor)) \,, $$ where $\chi(\,\cdot\,)$ is the alternating sum of cohomologies, $h^0(\,\cdot\,) - h^1(\,\cdot\,) + h^2(\,\cdot\,)$. The final term is often trivial to compute, through the Riemann-Roch theorem for surfaces. Hence if the Zariski decomposition is known and higher cohomologies are known to vanish for $\mathcal{O}(\lfloor P \rfloor)$, the zeroth cohomology of $\mathcal{O}(D)$ is easily obtainable.
In all the cases I have seen (in which the surface is usually toric or otherwise special), the higher cohomologies indeed always vanish for $\mathcal{O}(\lfloor P \rfloor)$. My question is:
Is it true in general that $h^q(S,\mathcal{O}(\lfloor P \rfloor)) = 0$ for $q>0\,$, or are there counterexamples?