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: Time Limit Exceeded)

N = int(input())
A = [0]*N
syogen = [0]*N

for i in range(N):
    A[i] = int(input())
    xy = [0]*A[i]
    for j in range(A[i]):
        xy[j] = [int(i) for i in input().split()]
    syogen[i] = xy

syojiki = [0]*N

truth = [7]*N

flag = True
flag2 = True

for i in range(N):
    truth[i] = 1
    for j in range(len(syogen[i])):
        truth[syogen[i][j][0]-1] = syogen[i][j][1]
    
    for j in range(N):
        if truth[j] == 1:
            for k in range(len(syogen[j])):
                if truth[syogen[j][k][0]-1] == 7:
                    truth[syogen[j][k][0]-1] = syogen[j][k][1]
                else:
                    if truth[syogen[j][k][0]-1] != syogen[j][k][1]:
                        flag = False
                        break

    if flag:
        tmp = 0
        for j in range(N):
            if truth[j] == 1:
                tmp += 1
        syojiki[i] = tmp

        while 7 in truth:
            for j in range(N):
                if truth[j] == 7:
                    truth[j] = 1
                    for k in range(len(syogen[j])):
                        if truth[syogen[j][k][0]-1] == 7:
                            truth[syogen[j][k][0]-1] = syogen[j][k][1]
                        else:
                            if truth[syogen[j][k][0]-1] != syogen[j][k][1]:
                                flag2 = False
                                break
                if not flag2:
                    break
        if flag2:
            tmp = 0
            for j in range(N):
                if truth[j] == 1:
                    tmp += 1
            syojiki[i] = tmp
        
    truth = [7]*N
    flag = True
    flag2 = True

print(max(syojiki))