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: Wrong Answer)

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

#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)>(b)?(b):(a))
#define SWAP(type, a, b) \
        do {type tmp = a; a = b; b = tmp;} while (0)


typedef long long ll;


int main(void){
    int N;
    int A[15];
    int x[15][15], y[15][15];
    int i, j;
    scanf("%d", &N);
    for(i=0;i<N;i++){
        scanf("%d", &A[i]);
        for(j=0;j<A[i];j++){
            scanf("%d %d", &x[i][j], &y[i][j]);
        }
    }

    int g[2][15];
    for(i=0;i<2;i++){
        for(j=0;j<15;j++){
            g[i][j]=0;
        }
    }

    for(i=0;i<N;i++){
        for(j=0;j<A[i];j++){
            if(y[i][j]==1){
                if(g[0][x[i][j]-1]==1){
                    g[0][i]=1;
                }else if(g[1][x[i][j]-1]==1){
                    g[1][i]=1;
                }else{
                    g[0][i]=1;
                    g[0][x[i][j]-1]=1;
                }
            }else{
                if(g[0][x[i][j]-1]==1){
                    g[1][i]=1;
                }else if(g[1][x[i][j]-1]==1){
                    g[0][i]=1;
                }else{
                    g[0][i]=1;
                    g[1][x[i][j]-1]=1;
                }
            }
        }
    }

    for(i=0;i<N;i++){
        if(g[0][i]==g[1][i]){
            printf("0");
        }
    }
    int cnt=0;
    for(i=0;i<N;i++){
        if(g[0][i]==1){
            cnt++;
        }
    }
    printf("%d", cnt);

    return 0;
}