You could look at surfaces with maximal Picard rank. A surface is said to have maximal Picard rank if $H^{1,1}(X) \cap H^1(X, \mathbb{R})$ is spanned by curve classes. So the Kahler cone is the same as the ample cone, and you can compute the intersection pairing on $H^{1,1}$ by just intersecting curves with each other.

Now, this raises the question of whether there is a surface of general type with maximal Picard rank. Some googling found me the following:

Section 3 of Quintic surfaces with maximum and other Picard numbers shows that the Fermat sextic $w^6+x^6+y^6+z^6=0$ has maximal Picard rank. Apparently, this was originally computed by Beauville.

Roulleau shows that there are infinitely many Fano surfaces with maximal Picard rank, and gives a basis for the Neron-Severi group. (Confusingly, the Fano surface is not a Fano variety, but rather is of general type.)

A warning: If you look at this paper or this one, they compute a lot of surfaces with maximal Picard. However, their method is to make a surface with an $A_n$ singularity and prove that its resolution has maximal Picard. Since this resolution has a $-2$ curve, it is automatically not of general type.

You could look at surfaces with maximal Picard rank. A surface is said to have maximal Picard rank if $H^{1,1}(X) \cap H^1(X, \mathbb{R})$ is spanned by curve classes. So the Kahler cone is the same as the ample cone, and you can compute the intersection pairing on $H^{1,1}$ by just intersecting curves with each other.

Now, this raises the question of whether there is a surface of general type with maximal Picard rank. Some googling found me the following:

Section 3 of Quintic surfaces with maximum and other Picard numbers shows that the Fermat sextic $w^6+x^6+y^6+z^6=0$ has maximal Picard rank. Apparently, this was originally computed by Beauville.

Roulleau shows that there are infinitely many Fano surfaces with maximal Picard rank, and gives a basis for the Neron-Severi group. (Confusingly, the Fano surface is not a Fano variety, but rather is of general type.)

A warning: If you look at this paper or this one, they compute a lot of surfaces with maximal Picard. However, their method is to make a surface with an $A_n$ singularity and prove that its resolution has maximal Picard. Since this resolution has a $-2$ curve, it is automatically not of general type. Sorry. As Donu says, I was confused. $K$ will not be ample, but it will be big, and the varieties are general type.

I was coming back to edit in one more idea -- if $C$ is the Klein quartic, then I think that $C \times C$ has maximal Picard rank. But Jason Starr's comment is simpler than any of my suggestions.