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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;

int main(){
        int n; scanf("%d", &n);

        vector<vector<P>> xy(n, vector<P>(0));
        for(int i=0; i<n; i++){
                int a; scanf("%d", &a);
                for(int j=0; j<a; j++){
                        int x, y;
                        scanf("%d %d", &x, &y);
                        xy[i].push_back(make_pair(x-1, y));
                }
        }

        int ans = 0;
        for(int i=0; i<(1<<n); i++){
                vector<bool> hone(n), evil(n);
                for(int j=0; j<n; j++) hone[j]=0, evil[j]=0;

                int tmp=0;
                for(int j=0; j<n; j++){
                        if((i>>j)&1){
                                tmp++;
                                if(evil[j]) tmp = -1e9;
                                else hone[j] = 1;
                                for(P pr: xy[j]){
                                        if(pr.second){
                                                if(!evil[pr.first]) hone[pr.first] = 1;
                                                else tmp = -1e9;
                                        }else{
                                                if(!hone[pr.first]) evil[pr.first] = 1;
                                                else tmp = -1e9;
                                        }
                                }
                        }else{
                                if(hone[j]) tmp=-1e9;
                                else evil[j] = 1;
                        }
                }
                ans = max(ans, tmp);
        }
        printf("%d\n", ans);
        return 0;
}