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;

int n, v;
int x[100][100], y[100][100], a[100], p[100];
int maxn;

int check() {
	for(int i = 1; i <= n; i++) {//i表示现在枚举到哪个人 
		if(!p[i]) continue;
		for(int j = 1; j <= a[i]; j++) {//j表示这个人说的每句话 (这个人每句话都是实话) 
			if(y[i][x[i][j]] != p[x[i][j]]) return 0;//y[i][x[i][j]]表示第i个人认为第x[i][j]个人的状态 //如果发现他有一句不是实话，则矛盾 
		}
	}
	return 1;
}

int dfs(int now, int current) {
	if(now == n + 1) {
		if(check()) {
			maxn = max(maxn, current);
		}
		return 0;
	}
	for(int l = 1; l >= 0; l--) {
		p[now] = l;
		dfs(now + 1, current + p[now]);
	}
	return 0;
}

int main() {
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
		for(int j = 1; j <= a[i]; j++) {
			cin >> x[i][j];
			cin >> y[i][x[i][j]]; 
		}
	}
	dfs(1, 0);
	cout << maxn << endl;
	return 0;
}