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>
int main(void){
int N;
  scanf("%d",&N);
  int A[N+1],x[N+1][211][2],y[N+1][211][2],mujunn=0,flag=0,max=0;
  for(int i=0;i<N;i++){
scanf("%d",&A[i]);
    for(int j=0;j<A[i];j++){
  scanf("%d %d",&x[i][j][0],&y[i][j][0]);
    }
  }
  int honest[N+1],c;
  for(int i=0;i<(1<<N);i++){///333
    c=0;
    mujunn=0;
    //printf("%dパターン目",i+1);
    for(int j=0;j<N;j++){
      if(i&(1<<j)){
      honest[c]=j+1;
        c++;
      //  printf("%dさん\t",j+1);
      }
    }
   // printf("合計%d人\n",c);
    for(int l=0;l<c;l++){  //正直者候補のうち誰に注目するか222
      if(mujunn==1){
        break;
      }
		for(int j=0;j<A[honest[l]-1];j++){  //証言を一つ一つ注目111
          if(mujunn==1)break;
          for(int k=0;k<c;k++){
            if(x[honest[l]-1][j][0]==honest[k]) {   //証言された人が正直者か否か
             										//for回し終えた時点でflag=0ならピエロ
              flag=1;}
          
        }
      
      if(!flag){
        if(y[honest[l]-1][j][0]==1){
          		mujunn=1;
              // printf("%dさんはピエロだが、%d(正直)と証言しているので矛盾\n",x[honest[l]-1][j][0],
                //        y[honest[l]-1][j][0]);
                                   }
      }
      else {
        if(y[honest[l]-1][j][0]==0){
          mujunn=1;
          // printf("%dさんは正直者だが、%d(ピエロ)と証言しているので矛盾\n",x[honest[l]-1][j][0],
            //            y[honest[l]-1][j][0]);
        }
      }
                flag=0;
       }//証言1つに関するfor文閉じ111
	  }//222   
      if(mujunn==0){          //矛盾テストクリアの時
        if(c>max){max=c;
        //printf("更新\n");
                 }
      }

      			
  }	///333
printf("%d",max);

  return 0;
}