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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
typedef long long ll;
#define sz(x) int(x.size())
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
int main(){
    int N;
    cin >> N;
    int A[N];
    int X[N][N];
    
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            X[i][j] = 1;
        }
        
    }
    
    for (int i = 0; i < N; i++)
    {
        cin >> A[i];
        int x,y;
        for (int j = 0; j < A[i]; j++)
        {
            cin >> x >> y;
            X[i][x-1] = y;
        }
        
    }

    int ans = N;
    while(true){
        int summin = 20;
        int cnt = -1;
        for (int i = 0; i < N; i++)
        {
            int sum = 0;
            for (int j = 0; j < N; j++)
            {
                sum += X[i][j];
            }
            if(sum < summin) {
                summin = sum;
                cnt = i;
                }
                
        }
        if(summin == N){
            cout << ans << endl;
            return 0;
        }
        else ans -= 1;
        for (int i = 0; i < N; i++)
        {
            if(X[cnt][i] == 0){
                X[cnt][i] = 1;
                X[i][cnt] = 1;
            }
        }
    }
  
}