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>

#define MAX_PEOPLE 15

int main(void)
{
    int i,j,k;
    int n; //人数
    int a; //１人あたりの証言の個数
    int x[MAX_PEOPLE][2];
    int y[MAX_PEOPLE][2];
    int honest[MAX_PEOPLE]={0};
    int unkind[MAX_PEOPLE]={0};
    int count=0;
    int sum_honest=0;
    int sum_unkind=0;

    //input
    scanf("%d", &n);
    for(i=0; i<n; i++){
        scanf("%d", &a);
        for(j=0; j<a; j++){
            scanf("%d %d", &x[i][j], &y[i][j]);
        }
    }

    for(i=0; i<n; i++){
        for(j=0; j<a; j++){
            if(y[i][j]==1){
                honest[i]=1;
                sum_honest++;
            }
            else{
                unkind[i]=1;
                sum_unkind++;
            }
        }
    }

    if(sum_honest == sum_unkind){
        count = 0;
    }
    else if(sum_honest==0){
        count = n/2;
    }
    else{
        count = sum_honest;
    }

    printf("%d\n", count);
}