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 N;//人数
typedef struct{
    int x,y;
}Vec;
typedef struct{
    int A;
    Vec *xy; 
}Hum;

int nofin(int array[]){//配列に絶対値を取ると1の要素がまだあると1を返す
    int i;
    for(i=0;i<N;i++) if(abs(array[i])==1) return 1;
    return 0;
}
int main(){
    scanf("%d",&N);
    Hum *human = malloc(sizeof(Hum)*N);
    int i,j,tmp1,tmp2;
    for(i=0;i<N;i++){
        scanf("%d",&human[i].A);
        human[i].xy=malloc(sizeof(Vec)*human[i].A);
        for(j=0;j<human[i].A;j++){
            scanf("%d %d",&tmp1,&tmp2);
            human[i].xy[j].x=tmp1-1;
            human[i].xy[j].y=(tmp2)?1:-1;//正直者という証言が1,嘘つきという証言が-1
        }
    }
    // i番目の人が正直者だと仮定したときの正直者の人数を考える
    int max=0,honest,*D,k;
    for(i=0;i<N;i++){
        int flag=1;
        //D[i]:=i番目の人が2,1,0,-1,-2でそれぞれ正直者で考慮済み,正直者,未確定,嘘つき,嘘つきで考慮済み
        D = calloc(N,sizeof(int));
        D[i]=1;
        while(nofin(D) && flag){
            for(j=0;j<N;j++) if (abs(D[j])==1) break;
            Hum *h = &human[j];
            for(k=0;k<h->A && flag;k++){
                int *d = &D[h->xy[k].x];
                int ass = h->xy[k].y*D[j];//正しい主張(正直者ならそのまま、嘘つきなら反転させた主張)
                if(*d==0){//注目している人物が言及している人物が未確定なとき
                    *d=ass;
                } else if (*d > 0){//正直者と確定しているとき
                    //嘘つきという主張の場合矛盾
                    if (ass < 0) flag=0;
                }else {//嘘つきと確定している場合
                    if (ass > 0) flag=0;
                }
            }
            D[j]*=2;
        }
        //D[]に1が消えるか矛盾が起こるとループを抜ける
        if(flag){//D[]から1が消えた
            int num=0;
            for(j=0;j<N;j++) {
                if(D[j]==2) num++;
            }
            if(max < num) max=num;
        }
        free(D);
    }
    printf("%d\n",max);
    return 0;
}