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>
using namespace std;
void bitrow(vector<int> &vec,int x,int n){
    for(int i=n-1;i>=0;i--){
        vec.at(i)=x/(int)pow(2,i);
        x%=(int)pow(2,i);
    }
}

int main(){
    int n,tmp1,tmp2;
    cin >> n;
    vector<int> a(0);
    vector<vector<int> > x(n,vector<int>(0));
    vector<vector<int> > y(n,vector<int>(0));
    for(int i=0;i<n;i++){
        cin >> tmp1;
        a.push_back(tmp1);
        for(int j=0;j<a.at(i);j++){
            cin >> tmp1 >> tmp2;
            x.at(i).push_back(tmp1-1);
            y.at(i).push_back(tmp2);
        }
    }
    vector<int> vec(n);
    int count=0;
    for(int i=0;i<pow(2,n);i++){
        int flag=1;
        bitrow(vec,i,n);
        vector<int> hantei(n);
        for(int j=0;j<n;j++)hantei.at(j)=vec.at(j);
        for(int j=0;j<n;j++){
            for(int k=0;k<a.at(j);k++){
                if(vec.at(j)==1){   //正直な場合
                    if(hantei.at(x.at(j).at(k))!=y.at(j).at(k)){
                        flag=0;
                        break;
                    }
                }
                else{   //嘘つきの場合
                    if(hantei.at(x.at(j).at(k))==y.at(j).at(k)){
                        flag=0;
                        break;
                    }
                }
            }
            if(flag==0)break;
        }
        if(flag==1){
            int sum=0;
            for(int j=0;j<n;j++){
                sum+=vec.at(j);
            }
            count=max(count,sum);
        }
    }
    cout << count << endl;
}