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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

#define N 15

int main() {
	static int aa[N], xx[N][N - 1], yy[N][N - 1], honest[N];
	int n, i, j, b, k, yes, ans;

	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", &aa[i]);
		for (j = 0; j < aa[i]; j++)
			scanf("%d%d", &xx[i][j], &yy[i][j]), xx[i][j]--;
	}
	ans = -1;
	for (b = 0; b < 1 << n; b++) {
		memset(honest, 0, n * sizeof *honest);
		for (i = 0; i < n; i++)
			if (b & 1 << i)
				honest[i] = 1;
		yes = 1;
		k = 0;
		for (i = 0; i < n; i++)
			if (honest[i]) {
				k++;
				for (j = 0; j < aa[i]; j++)
					if (honest[xx[i][j]] != yy[i][j])
						yes = 0;
			}
		if (yes && ans < k)
			ans = k;
	}
	printf("%d\n", ans);
	return 0;
}