Are there any relations between $H^n(S\setminus\{p\}, \mathcal{O})$ and $H^n(S, \mathcal{O})$ when $\{p\}$ is rational singular point? Let $S$ be a rational surface over $\mathbb{C}$, $\{p\} \subset S$ be a rational singular point.
My question is, 
are there any relations between $H^n(S\setminus\{p\}, \mathcal{O})$ and $H^n(S, \mathcal{O})$ in general? Or, how to compute $H^n(S\setminus\{p\})?$
I think, because $S$ is normal and codim$_{S}(\{p\}) = 2$,
$H^0(S\setminus\{p\}, \mathcal{O}) \simeq  H^0(S, \mathcal{O}) \simeq \mathbb{C}$. 
(Is it true?)
My motivation is as follow:
Suppose there exists a map $\pi: V \rightarrow S$ onto a rational surface S  which contracts some divisor $D$ of $V$ to $\{p\}$, i.e. which is an isomorphism at every point of $V \setminus D$ and such that $\pi(D)$ is a single point (rational singular point).
Then, I would like to know $H^n(V \setminus D, \mathcal{O})$.
 A: What you said is correct.
We have the long exact sequence in cohomology, let $U = X \setminus \lbrace p \rbrace$ 
$0 \rightarrow H^0_{p}(X, \mathcal{O}_X) \rightarrow H^0(X,\mathcal{O}_X) \rightarrow H^0(U,\mathcal{O}_X|_U) \rightarrow H^1_p(X,\mathcal{O}_X) \rightarrow H^1(X,\mathcal{O}_X) \rightarrow H^1(U,\mathcal{O}_U) \rightarrow H^2_p(X,\mathcal{O}_X) \rightarrow H^2(X,\mathcal{O}_X) \rightarrow H^2(U,\mathcal{O}_U) \rightarrow H^3_p(X,\mathcal{O}_X) \rightarrow 0$
Note that by excision we have $H^i_p(X,\mathcal{O}_X) = H^i_p(V, \mathcal{O}_V)$, where $V = spec(A), p = (m)$ is an open affine subset containing $p$. Since $V$ is affine, the cohomology with support in $p$ is $H^i_{m}(A) = H^i(A_m)$(see hartshorne page 217 exercise 3.4 (b)) and this is zero for $i < depth_m(A_m)$.
Now we know that $A$ is normal and hence serre's criterion says that $depth_m(A_m) \geq 2$, infact since $dim(A_m) = 2$, hence $depth_m(A_m) = 2$ and hence $H^1_m(A_m) = 0$ and hence $H^1_p(X, \mathcal{O}_X) = 0$. Plugging this we get
$0 \rightarrow H^1(X,\mathcal{O}_X) \rightarrow H^1(U,\mathcal{O}_U) \rightarrow H^2_p(X,\mathcal{O}_X) \rightarrow H^2(X,\mathcal{O}_X) \rightarrow H^2(U,\mathcal{O}_U) \rightarrow H^3_p(X,\mathcal{O}_X) \rightarrow 0$
Also note that $H^3_p(X,\mathcal{O}_X) = 0$ since $dim(X) < 3$(can be proved in a similar fashion as to the grothendiecke vanishing theorem). Thus we get
$0 \rightarrow H^1(X,\mathcal{O}_X) \rightarrow H^1(U,\mathcal{O}_U) \rightarrow H^2_p(X,\mathcal{O}_X) \rightarrow H^2(X,\mathcal{O}_X) \rightarrow H^2(U,\mathcal{O}_U) \rightarrow 0$
Note that we have not used the fact that $p$ is a "rational singularity". Now if the surface has only rational singularities then assuming $\tilde{X} \rightarrow X$ is a resolution, then $H^2(\tilde{X},\mathcal{O}_{\tilde{X}}) = H^0(\tilde{X}, \omega_{\tilde{X}}) = 0$ since $\tilde{X}$ is rational and arithmetic genus is a birational invariant. Now since $X$ has rational singularities we get that $H^2(X,\mathcal{O}_X) = H^2(\tilde{X}, \mathcal{O}_{\tilde{X}}) = 0$. Thus we get 
$H^2(U,\mathcal{O}_U) = 0$
and
$0 \rightarrow H^1(X,\mathcal{O}_X) \rightarrow H^1(U,\mathcal{O}_U) \rightarrow  H^2_p(X,\mathcal{O}_X) \rightarrow 0$.
Ideally, $H^2_p(X,\mathcal{O}_X)$ can be non-zero.
