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)

# 入力の受取
n = int(input())
a = [0] * (n+1)
x = [[0] * (n+1) for _ in range(n+1)]
y = [[0] * (n+1) for _ in range(n+1)]

for i in range(1,n+1):
    a[i] = int(input())
    for j in range(1,a[i]+1):
        x[i][j], y[i][j] = map(int, input().split())

# ビット全探索用のシーケンスを生成
import itertools
seq = [0, 1]
arr = itertools.product(seq, repeat=(n))

# ビット全探索
# 証言と整合する場合は，その際の人数をリストに格納
ans = [0]
for v in arr:
    v = [0] + list(v)
    
    flag = True
    # 証言と整合しない場合，フラグをfalseにする
    for i in range(1, n+1):
        if v[i] == 1:
            for k in range(1,a[i]+1):
                if y[i][k] == 1 and v[x[i][k]] == 0:
                    flag = False
                if y[i][k] == 0 and v[x[i][k]] == 1:
                    flag = False
                          
    # 証言と整合した場合
    if flag:
        ans.append(sum(v))


# 証言と整合したケースの人数の最大値を出力
print(max(ans))