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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

typedef struct node{
    int a;
    int y[15];
}node;

void f(node* n){
    n->a = -1;
    for(int i=0; i<15; i++){
        n->y[i] = -1;
    }
}

int main(){
    int n;
    scanf("%d",&n);
    node a[n];
    for(int i=0; i<n; i++){
        f(&a[i]);
    }
    int x[n][15], y[n][15];
    for(int i=0; i<n; i++){
        scanf("%d", &(a[i].a));
        for(int j=0; j<a[i].a; j++){
            scanf("%d%d", &x[i][j], &y[i][j]);
        }
    }
    for(int i=0; i<n; i++){
        for(int j=0; j<a[i].a; j++){
            a[i].y[x[i][j] - 1] = y[i][j];
        }
    }
    int bi = pow(2, n) - 1;
    int max = 0;
    for(int i=bi; i>=0; i--){
        int k=i;
        int target = 0;
        int flag = 1;
        int count = 0;
        while(k > 0){
            if(k%2 == 1){
                count++;
                for(int j=0; j<n; j++){
                    if(a[target].y[j] == -1){
                        continue;
                    }else{
                        int buf = i;
                        for(int l=0; l<j; l++){
                            buf = buf/2;
                        }
                        if(buf % 2 != a[target].y[j]){
                            flag = 0;
                        }
                    }
                }
            }
            k = k/2;
            target++;
        }
        if(flag){
            //printf("%d\n", i);
            if(max < count){
                max = count;
            }
        }
    }
    printf("%d", max);

    return 0;
}