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>
#include<string.h>
#include <time.h>
// #define long long llint
long long n2place(long long a, long long b){
    return b==1?a%2:n2place(a/2,b-1);
}
typedef struct{
        long long aa;
        long long bb;
    }s;
int cmp( const void *p, const void *q ) {
    return ((s*)p)->bb - ((s*)q)->bb;
}
long long standnum2(long long a){
    long long count=1,i,b;
    b=a;
    while(1){
        if(b/2==0) break;
        count++;
        b/=2;
    }
    long long stand=0;
    for(i=0;i<count;i++){
        if(n2place(a,i+1)==1) stand++;
    }
    return stand;
}

int main(void){
    long long i,j,k,n,count,honest=0;
    scanf("%lld",&n);
    long long a[n],x[250],y[250],suma[n+1],judge[n];
    suma[0]=0;
    for(i=0;i<n;i++){
        scanf("%lld",&a[i]);
        for(j=0;j<a[i];j++){
            scanf("%lld",&x[suma[i]+j]);
            scanf("%lld",&y[suma[i]+j]);
        }
        suma[i+1]=suma[i]+a[i];
    }
    // for(j=0;j<n+1;j++){
    //         printf("%lld ",suma[j]);
    //     }
    count = pow(2,n);
    s number[count];
    for(i=0;i<count;i++){
        number[i].aa=i;
    }
    for(i=0;i<count;i++){
        number[i].bb=standnum2(i);
    }
    qsort(number,count,sizeof(s),cmp);
    // for(i=0;i<count;i++){
    //     printf("%lld ",number[i].bb);
    // }

    for(i=count;i>0;i--){
        for(j=0;j<n;j++){
            judge[j] = n2place(number[i-1].aa,j+1);
            // if(n2place(count,j+1) == 0){
            //     judge[j]=0;
            //     for(k=0;)
            // }else{
            //     judge[j]=1;
            // }
        }
        // for(j=0;j<n;j++){
        //     printf("%lld ",judge[j]);
        // }
        // printf("\n");
        for(j=0;j<n;j++){
            if(judge[j]==0) continue;
            for(k=suma[j];k<suma[j+1];k++){
                if(y[k]!=judge[x[k]-1]) break;
            }
            if(k!=suma[j+1]) break;
        }
        if(j==n){
            for(j=0;j<n;j++){
                if(judge[j]==1) honest++;
            }
            break;
        }
    }
    if(i==0) honest=i;

    printf("%lld\n",honest);
    return 0;
}