To get AC you should not make the sum of all the types of nearest brangches but the "or" in pascal and "|" in c++. examle:
1+5+2=8 but you get WA
1|5|2=7 in c++ to get AC
1 or 5 or 2 =7 in pascal to get AC
    I don't know why, but this is the way to get accepted. Can anyone explain why? (the problem text seems ambiguos to me).

You must find sum of (!) different branches for each cross-road.

In your test:
1 = 1
2 = 2
5 = 1+4

So, there are 3 different branches(1,2,4) and sum = 7

Thanks a lot! To use or is more simple then solve that problem without it! )