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: Wrong Answer)

import itertools
n = int(input())
l = []
flag = True
for i in range(n):
    a = int(input())
    tmp = []
    for j in range(a):
        a1, a2 = map(int, input().split())
        if a2 == 0:
            flag = False
        l.append([i, a1, a2])

if flag:  # 全員正直の場合
    print(n)
else:
    for i in range(1, n + 1):
        # jが嘘つきと仮定する
        for j in itertools.combinations(range(n), i):
            mujun = False
            # tmp2に真偽入っている
            tmp2 = [0 if i in j else 1 for i in range(n)]
            # print(tmp2)
            for k in l:
                # print(k)
                # 嘘つきの場合
                if tmp2[k[0]] == 0:
                    if tmp2[k[1] - 1] == k[2]:
                        mujun = True
                        break  # 破綻
                # 正直者の場合
                else:
                    if tmp2[k[1] - 1] != k[2]:
                        mujun = True
                        break
            if mujun == False:
                # print('in')
                print(n - i)
                exit()
    print(0)

    # print(l)