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: Wrong Answer)

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



int main() {

	int N;
	int A[15];
	int X[15][15];
	int Y[15][15];
	int MaxHonest = 0;
	int Honest = 0;
	int Flag = 0;
	
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		scanf("%d", &A[i]);
		for (int j = 0; j < A[i]; j++) {
			scanf("%d %d", &X[i][j], &Y[i][j]);
		}
	}

	for (int i = 0; i < (1 << N); i++) {
		for (int j = 0; j < N; j++) {
			if (i & (1 << j)) {
				for (int k = 0; k < A[j]; k++) {
					if (Y[j][k] == 0) {
						if (i & (1 << X[j][k] - 1)) {
							Flag = 1;
							break;
						}
					}
					else {
						if (!(i & (1 << X[j][k] - 1))) {
							Flag = 1;
							break;
						}
					}
				}
			}
			else {
				continue;
			}
			if (Flag == 0) {
				Honest += 1;
				if (MaxHonest < Honest) {
					MaxHonest = Honest;
				}
			}
			else {
				Flag = 0;
			}
		}
		Honest = 0;
	}

	printf("%d", MaxHonest);
	
	return 0;
}