Not really.

In the example given (20 cc to pour), the result would be

A B C D
1 7 3 5 4
2 4 1 3 2
3 9 10 10 10

Before pouring, I had 7+3+5+4 = 19 cubic centimeters (cc)
After pouring I can have, at most 19+20 = 39 cc
Because I start with the glass that is more empty, I can only pour 2 cc
in my first glass, giving a total of 7+2 = 9 cc

Another example, say I only have 10 cc to pour. Then,

A B C D
1 7 3 5 4
2 4 1 3 2
3 7 10 5 7

I start pouring in the glass with 3 cc, giving 3+7 = 10 cc. The other 3
go to the glass having 4 cc giving 4+3 = 7 cc. The other two glasses
are unchanged.

I hope it is clearer.