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>
#include <math.h>

using namespace std;

long long inf=1000000007;

int main(){
	int n;
	cin>>n;
	vector<vector<int>> tes(n,vector<int>(n+1,0));
	for(int i=0;i<n;i++){
		int cnt;
		cin>>cnt;
		for(int j=0;j<cnt;j++){
			int a,b;
			cin>>a>>b;
			if(b==0) b=-1;
			tes.at(i).at(a)=b;
		}
	}

	int ans=0;

	for(int i=0;i<(1<<15);i++){
		bitset<15> bits=i;
		map<int,bool> hoge;
		bool flag=true;
		int pre=0;

		for(int j=0;j<n;j++){
			if(bits.test(j)) pre++;
		}

		for(int j=0;j<n;j++){
			if(!bits.test(j)) continue;

			for(int k=1;k<=n;k++){
				if(tes.at(j).at(k)!=0){
					if(!hoge.count(k)) hoge[k]=tes.at(j).at(k);
					else if(hoge.at(k)!=tes.at(j).at(k)) flag=false;

					if(tes.at(j).at(k)==1 && tes.at(k-1).at(j+1)==-1) flag=false;
					if(tes.at(j).at(k)!=bits.test(k-1)) flag=false;
				}
			}
		}

		if(flag) ans=max(ans,pre);
	}

	cout<<ans<<endl;

	return 0;
}