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: Accepted)

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

typedef struct {
    int x,y;
}Vec;

typedef struct{
    int A;
    Vec *Tes;
} Human;

int main(){
    int N,i,j;
    // 入力---------
    scanf("%d",&N);
    Human *Hum = malloc(sizeof(Human)*N);
    for(i=0;i<N;i++){
        scanf("%d",&Hum[i].A);
        Hum[i].Tes=malloc(sizeof(Vec)*Hum[i].A);
        for(j=0;j<Hum[i].A;j++) scanf("%d %d",&Hum[i].Tes[j].x,&Hum[i].Tes[j].y);
    }
    int bit,max=0;
    for(bit=0;bit<(1<<N);bit++){
        int flag=0,num=0;
        for(i=0;i<N && (!flag);i++){
            if((bit>>i)&1) {//i桁目のビットが立っていたらHum[i]が正直者
                num++;
                for(j=0;j<Hum[i].A;j++){//証言に合致するか検証
                    int index = Hum[i].Tes[j].x-1;
                    //証言に合致しないものを見つけたら旗を立てる
                    if(((bit>>index)&1)!=Hum[i].Tes[j].y) {
                        flag=1;
                        break;
                    }
                }
            }
        }
        if(!flag && max < num) max = num;
    }
    printf("%d\n",max);
}