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>
int x[16][16], y[16][16];
int pow2[16]={1,2,4,8,16,32,64,128,256,512,1024,
2048,4096,8192,16384,32768};
int binary[16];
int work1(int n)
{
	int i = 1;
	while(n)
	{
		binary[i] = n%2;
		i++;
		n /= 2;
	}
	return i;
}
int work2(int t)
{
	int cnt = 0;
	int flag = 0;          //flag=1表示有冲突了 
	for(int i=t-1;i>=1;i--)
	{
		if(binary[i]==1)    //此次讨论情况中假设了第i个人诚实
		{
			cnt++;
			for(int j=1;x[i][j]!=0;j++)  //则他说的话都是真的 
			{
				if(y[i][j]==1)    //第i个人说x[i][j]诚实 
				{
					if(binary[x[i][j]]!=1)      //看看是不是真诚实 
					{
						flag = 1;
						break;
					}
				}
				else
				{
					if(binary[x[i][j]]!=0)
					{
						flag = 1;
						break;
					}
				}
				if(flag) break;
			}
		}
		else
		{
			for(int j=1;x[i][j]!=0;j++)
			{
				if(y[i][j]==1)
				{
					if(binary[x[i][j]]==1)
					{
						flag = 1;
						break;
					}
				}
				else
				{
					if(binary[x[i][j]]==0)
					{
						flag = 1;
						break;
					}
				}
				if(flag) break;
			}
		}
		if(flag) break;
	}
	if(!flag) return cnt;
	else return 0;
}
int main()
{
	int n, ans = -1;
	scanf("%d", &n);
	for(int i=1;i<=n;i++)
	{
		int t;
		scanf("%d", &t);
		for(int j=1;j<=t;j++)
		{
			scanf("%d %d", &x[i][j], &y[i][j]);
		}
	}
	for(int i=1;i<=pow2[n];i++)
	{
		int t = work1(i);     //把i换成2进制数
		int temp = work2(t);
		if(temp>ans) ans = temp;
	}
	printf("%d\n", ans);
	return 0;
}