Since $f(2011) = 1$, we have:
$$m\sin(2011\pi + \alpha_1) + n\cos(2011\pi + \alpha_2) = 1.$$

Using the properties of sine and cosine concerning odd and even multiples of $\pi$, we have:
$$\sin(2011\pi + \alpha_1) = \sin(\alpha_1)$$
and
$$\cos(2011\pi + \alpha_2) = -\cos(\alpha_2).$$
This simplification arises because $2011\pi$ is an odd multiple of $\pi$ which makes sine function remain the same due to its periodicity, while cosine changes sign.

Hence, we can rewrite the equation as:
$$-msin(\alpha_1) - n\cos(\alpha_2) = 1,$$
which leads to:
$$msin(\alpha_1) + n\cos(\alpha_2) = -1$$

Now, let's find $f(2012)$ using an analogous argument:
$$f(2012) = m\sin(2012\pi + \alpha_1) + n\cos(2012\pi + \alpha_2).$$

Since $2012\pi$ is an even multiple of $\pi$, both $\sin$ and $\cos$ functions will not change sign with respect to their respective offsets, thus we have:
$$\sin(2012\pi + \alpha_1) = \sin(\alpha_1)$$
and
$$\cos(2012\pi + \alpha_2) = \cos(\alpha_2).$$

Substitute these values into the expression for $f(2012)$:
$$f(2012) = m\sin(\alpha_1) + n\cos(\alpha_2) = -1.$$

Therefore, the answer is $\boxed{-1}$.

Since $f(2011) = 1$, we have:
$$m\sin(2011\pi + \alpha_1) + n\cos(2011\pi + \alpha_2) = 1,$$
which simplifies to $-m\sin\alpha_1 - n\cos\alpha_2 = 1$. Therefore, we can deduce that $m\sin\alpha_1 + n\cos\alpha_2 = -1$.

Thus, $f(2012) = m\sin(2012\pi + \alpha_1) + n\cos(2012\pi + \alpha_2)$
$$= m\sin\alpha_1 + n\cos\alpha_2 = -1.$$

Therefore, the answer is $\boxed{-1}$.