is it possible to explicitly parametrise all the tstructures on the derived category of finitely generated abelian groups?

I guess the answer is the following. Take arbitrary subset $S$ of the set of all prime numbers. Let $A_S = <Z,\{Z/pZ\}_{p \not\in S},\{Z/pZ[1]\}_{p \in S}>$. Then $A_S$ is a heart of the tstructure which is obtained by a simple tilting from the standard one. The claim is that any bounded tstructure is a shift of one of those. The main part of the proof is to show that all objects $Z/pZ$ and $Z$ are pure in any tstructure. For this use the fact that $Z$modules have homological dimension 1 and that the objects from the heart of a tstructure don't have negative $Ext$'s between. EDIT: Let us show that $Z/pZ$ is pure. By shifting the tstructure we can assume that $Z/pZ \in D^{\le 0}$, but $Z/\pZ \not\in D^{\le 1}$. Then consider the triangle $$ Z/pZ \to H^0(Z/pZ) \to \tau^{\le 1}(Z/pZ)[1]. $$ Recall that any object in $D^b(Z)$ is a direct sum of $Z[i]$ and $Z/q^r Z[j]$ with $q$ prime. If there is a summand $M$ in $H^0(Z/pZ)$ such that $Hom(Z/pZ,M) = 0$ then $M$ is also a summand of $\tau^{\le 1}(Z/pZ)[1]$, hence $M \in D^{\le 1}[1] = D^{\le 2}$ and simultaneously $M \in D^0$ (since both subcategories are closed w.r.t. taking direct summands), so there is a nontrivial $Hom$ from $D^{\le 2}$ to $D^0$ which is impossible. So, all summands in $H^0(Z/pZ)$ should have a nontrivial $Hom$ from $Z/pZ$ into them. Thus we can only have $Z[1]$, $Z/p^rZ$ or $Z/p^rZ[1]$. Assume $H^0(Z/pZ) = Z^a[1] + \oplus_i Z/p^{r_i}Z + \oplus_j Z/p^{s_j}Z[1]$ and the above triangle has form $$ Z/pZ \to Z^a[1] + \oplus_i Z/p^{r_i}Z + \oplus_j Z/p^{s_j}Z[1] \to \tau^{\le 1}(Z/pZ)[1]. $$ Now if we look at it in the standard tstructure we shall see that the third term has a $p$torsion cohomology in degree 0 and a cohomology in degree $1$ which is a sum of $Z^a$ and a $p$torsion. Since any complex in $D(Z)$ is a sum of its cohomology, we conclude that $\tau^{\le 1}(Z/pZ)$ is a sum of $Z^a$, $p$torsion and $p$torsion shifted by $1$. Once again, we cannot have same summands in $D^0$ and $D^{\le 1}$, hence the only possibility is that $H^0(Z/pZ) = Z/pZ$, so it is pure. A similar argument shows that $Z$ is also pure. 


The question and Sasha's answer can be generalized to a Noetherian ring $R$. The parametrization is in terms of functions from the integers to a specialization closed subset of $Spec R$. I do not have access to precise references at the moment, but see the paper 'Invariants of tstructures and classiﬁcation of nullity classes' by Don Stanley and further references therein. 

