Those are just ordinary bridge rectifiers. The choice of filter capacitor has little to do with the rectifier, and everything to do with the load you need to drive.
The capacitor gets charged in short bursts at the peaks of the AC waveform (at twice the line frequency for a bridge rectifier), and it supplies the load current during the intervals in between. Therefore, you can model the capacitor voltage as a sawtooth waveform. This waveform is called the "ripple voltage". The peak-to-peak voltage of this waveform gets larger with increasing load current, and it gets smaller if you increase the capacitance.
So, the two things you need to know are the average load current, and the maximum ripple you can tolerate. The maximum voltage is determined by the transformer, and the minimum voltage is determined usually by whatever regulator you're using.
You can derive the formula from the basic equation for a capacitor:
$$Q = CV$$
Q is charge, C is capaitance, and V is voltage. Since charge is just current × time, you can write this as:
$$It = CV$$
Solving for C gives:
$$C = \frac{It}{V}$$
This tells us that for a given load current I, a charging period t (e.g., \$\frac{1}{120}\$ s @ 60 Hz) and a peak-to-peak ripple voltage V, you need a capacitor with the value C.
For example, suppose you have a 8 VAC (RMS) transformer and you want to produce 5 V @ 1 A using a 7805 regulator, which needs an input of at least 8 VDC.
The peak voltage on the capacitor will be 8 VAC × 1.414, minus two diode drops in the rectifier, or about 9.9 VDC. Since the regulator needs 8 VDC, we can only allow 1.9 V of ripple. Again, assuming 60 Hz, the charging period is 8.33 ms. The formula gives us
$$C = \frac{It}{V} = \frac{1\text{ A} \cdot 8.33\text{ ms}}{1.9\text{ V}} = 4.4\text{ mF}$$