Solve in Python:
There are N people numbered 1 to N. Each of them is either an honest person whose testimonies are always correct or an unkind person whose testimonies may be correct or not.
Person i gives A_i testimonies. The j-th testimony by Person i is represented by two integers x_{ij} and y_{ij}. If y_{ij} = 1, the testimony says Person x_{ij} is honest; if y_{ij} = 0, it says Person x_{ij} is unkind.
How many honest persons can be among those N people at most?

-----Constraints-----
 - All values in input are integers.
 - 1 \leq N \leq 15
 - 0 \leq A_i \leq N - 1
 - 1 \leq x_{ij} \leq N
 - x_{ij} \neq i
 - x_{ij_1} \neq x_{ij_2} (j_1 \neq j_2)
 - y_{ij} = 0, 1

-----Input-----
Input is given from Standard Input in the following format:
N
A_1
x_{11} y_{11}
x_{12} y_{12}
:
x_{1A_1} y_{1A_1}
A_2
x_{21} y_{21}
x_{22} y_{22}
:
x_{2A_2} y_{2A_2}
:
A_N
x_{N1} y_{N1}
x_{N2} y_{N2}
:
x_{NA_N} y_{NA_N}

-----Output-----
Print the maximum possible number of honest persons among the N people.

-----Sample Input-----
3
1
2 1
1
1 1
1
2 0

-----Sample Output-----
2

If Person 1 and Person 2 are honest and Person 3 is unkind, we have two honest persons without inconsistencies, which is the maximum possible number of honest persons.
n=int(input())
a=[]
l=[]
for i in range(n):     
      A=int(input())
      L=[list(map(int,input().split())) for _ in range(A)]
      a.append(A)
      l.append(L)
ans=0
for i in range(2**n):
      b=[0]*n
      for j in range(n):
            if (i>>j)&1:
                  b[j]=1
      for k in range(n):
            for h in range(a[k]):
                  hito=l[k][h][0]-1
                  singi=l[k][h][1]
                  if b[k]==1 and b[hito]!=singi:
                        break      
            else:
                  continue
            break
      else:
            ans=max(ans,sum(b))

print(ans)