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>
#include <stdlib.h>

int x[16][16];
int y[16][16];
int A[16]={};

int judge(int bit, int N, int A[])
{
    int mask;
    for(int i=0; i<N; i++) { //N人について正直か不親切かを判定する
        if((bit>>i)==0) continue;

        for(int j=0; j<A[i]; j++) {
            mask=(1<<x[i][j]);
            if(y[i][j]==1 && !(bit & mask)) {
                return 0; //正直者の証言と仮定したものが等しいか比べる。
            } else if(y[i][j]==0 && (bit & mask)==mask) return 0; //bit & maskをしたとき、maskで残した部分のフラグが立っていなかったら0を返すことになるのでif文の条件を満たさなくなる。
        }
    }
    return 1;
}

int main()
{
    
    int N;
    int cnt=0;
    scanf("%d", &N);

    for(int i=0; i<N; i++) { //人数、証言の数、どの人が正直or不親切かを読み取る
        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<N; i++) {
        int bit=i;
        if(judge(i, N, A)) {
            for(int j=0; j<N; j++) {
                if(bit & 1<<j) cnt++;
            }
        }
    }

    printf("%d", cnt);

}