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)

import sys
sys.setrecursionlimit(100000)
N = int(input().strip())
shogen = [[-1 for _ in range(N)] for j in range(N)]
for i in range(N):
    size = int(input().strip())
    for j in range(size):
        k, tf = map(int, input().strip().split())
        k -= 1
        shogen[i][k] = tf

def dfs(bp, k):
    if k == N:
        # 矛盾なら0 そうでないならbp = Trueの数を返す
        for i in range(N):
            if not bp[i]:
                continue
            for j in range(N):
                if shogen[i][j] == 1 and not(bp[j]):
                    return 0
                if shogen[i][j] == 0 and bp[j]:
                    return 0
        cnt = 0
        for i in range(N):
            if bp[i]:
                cnt += 1
        return cnt
    """
    #　矛盾チェック
    mujun = False
    for j in range(N):
        if j == k:
            for l in range(N):s
                if shogen[k][l] == 0
        else:
            if shogen[j][k] == 0:
                mujun = True
                break

    """
    bp[k] = True
    tmp1 = dfs(bp, k + 1)
    bp[k] = False
    tmp2 = dfs(bp, k + 1)

    return max(tmp1, tmp2)

tmp_bp = [False] * N
ans = dfs(tmp_bp, 0)
print(ans)