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 <stdlib.h>

int main()
{
	int i, j, N, A[16], x[16][16], y[16][16];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) {
		scanf("%d", &(A[i]));
		for (j = 1; j <= A[i]; j++) scanf("%d %d", &(x[i][j]), &(y[i][j]));
	}
	
	int honest[16] = {}, max = 0, count;
	while (1) {
		for (i = 1; i <= N; i++) if (honest[i] == 0) break;
		if (i > N) break;
		for (honest[i--] = 1; i >= 1; i--) honest[i] = 0;
		
		for (i = 1; i <= N; i++) {
			if (honest[i] == 0) continue;
			for (j = 1; j <= A[i]; j++) {
				if (honest[x[i][j]] != y[i][j]) break;
			}
			if (j <= A[i]) break;
		}
		
		if (i <= N) continue;
		count = 0;
		for (i = 1; i <= N; i++) count += honest[i];
		if (count > max) max = count;
	}
	
	printf("%d\n", max);
	fflush(stdout);
	return 0;
}