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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        //g[i][j]:i番目の人がj番目の人の事を何と言っているのか(正直者=1,不親切=0,情報なし=-1)
        int[][] g = new int[N][N];
        int ans = 0;

        //全て情報なしに初期化
        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                g[i][j] = -1; 
            }
        }
        //入力
        for(int i = 0; i < N; i++){
            int A = sc.nextInt();
            for(int j = 0; j < A; j++){
                int x = sc.nextInt();
                int y = sc.nextInt();
                g[i][x-1] = y; //0番目から14番目と番号を振る
            }
        }

        //2^N通りをbit全探索, iのbit列で正直者と不親切な人を決定する
        //例：0010:4人中2番目の人だけが正直者
        for(int i = 0; i < (1<<N); i++){
            int[] d = new int[N];
            for(int j = 0; j < N; j++){
                if((1 & i>>j) == 1){ //iのjbit目は1か?(右から0bit目,1bit目, ...)
                    d[j] = 1; //個々のiの場合において,j番目の人が正直者:d[j] = 1
                }
            }

            //証言が矛盾していないかをO(N^2)で調べる
            boolean ok = true;
            for(int j = 0; j < N; j++){
                if(d[j] == 1){ //正直者の言うことだけ調べる
                    for(int k = 0; k < N; k++){
                        if(g[j][k] == -1) continue; //情報なしはスキップ
                        if(g[j][k] != d[k]) ok = false;
                        //正直者のk番目の人についての証言とその真偽が一致しなかったのでこのiは矛盾
                    }
                }
            }

            //bit列iに1がいくつあるか数える(正直者の人数)
            int popcount = 0;
            for(int l = 0; l < N; l++){
                if((1 & i>>l) == 1) popcount++;
            }
            if(ok) ans = Math.max(ans, popcount);
        }

        System.out.println(ans);
    }
}