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<string.h>
#include<stdint.h>
#include<stdbool.h>
#include<tgmath.h>
#include<sys/queue.h>
#include<time.h>
#define ll long long
#define min(a, b) ((a) > (b) ? (b) : (a))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define miin(a, b, c) min((a), min((b), (c)))
#define maax(a, b, c) max((a), max((b), (c)))
#define mod 1000000007
#define INF 1001001001

int n, a[15], x[15][15][2], ans = 0;

main(){
    scanf("%d", &n);
    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], &x[i][j][1]);
            x[i][j][0]--;
        }
    }
    for(int i = 0; i< (1<<n); i++){
        int cnt = 0;
        for(int j=0; j<n; j++){
            if(i & (1<<j)){
                for(int k=0; k<a[j]; k++){
                    if(((i>>x[j][k][0])&1) != x[j][k][1]){
                        cnt = -INF;
                        break;
                    }
                }
                cnt++;
            }
        }
        ans = max(ans, cnt);
    }
    printf("%d\n", ans);
    return 0;
}