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 <iostream>
#include <utility>
#include <map>
#include <vector>
#include <algorithm>
#include <queue>
#include <math.h>
#include <stack>
#include <set>
#include <deque>

using namespace std;
typedef long long ll;
ll mod = 1e9+7;
#define rep(i,n) for(int i = 0; i < (n); ++i)

struct edge {
    int to;
    int cost;
};

int dfs ( int s, vector<vector<edge>> &g) {
    int cnt = 0;
    for ( auto u : g[s] ) {
        for ( auto v : g[u.to] ) {
            if ( s == v.to ) {
                if ( u.cost && v.cost ) {
                    cnt += 2;
                }
                else if ( !u.cost && !v.cost ) {
                    ++cnt;
                }
            }
        }
    }
    return cnt;
}

int main(){
    int n;
    cin >> n;
    vector<vector<edge>> g(n);
    rep(i, n) {
        int a;
        cin >> a;
        rep(j, a) {
            int x, y;
            cin >> x >> y;
            --x;
            edge e = {x, y};
            g[i].push_back(e);
        }
    }
    int ans = 0;
    rep(i, n) {
        ans += dfs(i, g);
    }
    cout << ans / 2 << endl;
    return 0;
}