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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int counter(int z){
    if(z==0) return 0;
    else return counter(z>>1)+(z&1);
}

int main(void)
{
    int n=0,max=0,count=0,flag=1;
    long bit=1;
    int a[20]={},x[20][20]={},y[20][20]={};
    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(long i=1;i<(1<<n);i++){
        flag=1;
        for(int j=0;j<n;j++){
            for(int k=0;k<a[j];k++){
                if(((i>>(x[j][k]-1))&1)^y[j][k]){
                    flag=0;
                    break;
                }
            }
        }
        // 矛盾がなかった時
        if(flag==1){
            count=counter(i);
            if(max<count) max=count;
        }
    }
    printf("%d",max);
    return 0;
}

void zeros(int *a){
    for(int i=0;i<20;i++) a[i]=0;
}