Problem:
Score : 300 points

Problem Statement

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 1

3
1
2 1
1
1 1
1
2 0

Sample Output 1

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.

Sample Input 2

3
2
2 1
3 0
2
3 1
1 0
2
1 1
2 0

Sample Output 2

0

Assuming that one or more of them are honest immediately leads to a contradiction.

Sample Input 3

2
1
2 0
1
1 0

Sample Output 3

1

Solution in Python (3.4.3) (judge verdict: Accepted)

#abc 147 C
import itertools
n = int(input())
word_g = []
for i in range(n):
    m = int(input())
    part_word = []
    for j in range(m):
        part_word.append(list(map(int, input().split())))
    word_g.append(part_word)
#print(word)       
state = list(itertools.product([0,1], repeat = n))  

def abc147c(n, word, i):
    word_i = []
    
    for j1 in range(n):
        if i[j1]  == 0:
            word_i.append([])
        else:
        	word_i.append(word[j1])
          
    
    check_mujun = [[] for i in range(n)]
    
    for j in word_i:
        for k in j:
            check_mujun[k[0] -1].append(k[1])
    
    
    counter = 0
    
    for x, j in enumerate(check_mujun):
        
        if len(set(j)) == 0  :
            counter += i[x]
            
        elif len(set(j)) == 1 and sum(set(j)) == i[x]:
            counter += i[x]
            
        else:
            return -1
            break
    return counter

output = 0 
for i in state:
    output = max(output, abc147c(n,word_g,list(i) ) )
    #print(i, abc147c(n,word_g,i))
print(output)