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<math.h>
int larger(int a,int b){
 return (a<=b)? b:a;}

int main(){
 int n;
 scanf("%d",&n);
 int a[n],x[n][n-1],y[n][n-1];
 int l,m;
 for(l=0;l<n;l++){
  scanf("%d ",&a[l]);
  for(m=0;m<a[l];m++){
   scanf("%d%d ",&x[l][m],&y[l][m]);
  }}
  
  int b,sb,sc,cnt=0,ans=0,ok=1;
  for(b=0;b<(1<<n);b++){ //正直者の数を仮定
   cnt=0;ok=1;
   for(sb=0;sb<n&ok;sb++){
    if(b&(1<<sb)){//正直者とされたsb人目の証言について吟味
     //printf("b=%d a[sb]=%d\n",b,a[sb]);
     cnt++;
     for(sc=0;sc<a[sb];sc++){//sc個目の証言について吟味
      //printf("b=%d sb=%d sc=%d cnt=%d\n",b,sb,sc,cnt);
     //printf("x[sb][sc]=%d y[sb][sc]=%d\n",x[sb][sc],y[sb][sc]);
      if((b&(1<<(x[sb][sc]-1)))!=(y[sb][sc]<<(x[sb][sc]-1))){
       ok=0;/*puts("NG");*/break;}
     }}}
  
  if(ok==1){
   //printf("b=%d cnt=%d\n",b,cnt);
   ans=larger(ans,cnt);} }
  printf("%d\n",ans);
    
return 0;}