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>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;

int main()
{
  int n; cin >> n;
  vector<int> a(n, 0);
  vector<vector<ii>> xy(n, vector<ii>());
  for (int i=0; i<n; i++) {
    cin >> a[i];
    for (int j=0; j<a[i]; j++) {
      int x, y; cin >> x >> y;
      x--;
      xy[i].push_back(make_pair(x, y));
    }
  }

  int ans = 0;
  for (int i=0; i<n; i++) {
    vector<int> d(n, -1);
    d[i] = 1;
    for (auto v: xy[i]) {
      d[v.first] = v.second;
    }
    bool flg = true;
    for (int k=0; k<2; k++) {
    for (int j=0; j<n; j++) {
      if (i==j) continue;

      for (auto v: xy[j]) {
        if (d[j] == 1 && d[v.first] != v.second) {
          flg = false;
          break;
        }
        if (d[j] == 0 && d[v.first] == v.second) {
          flg = false;
          break;
        }
        if (d[j] == 1 && d[v.first] == -1) {
          d[v.first] = v.second;
        }
      }
    }
    }
    int t = 0;
    if (flg) {
      for (auto v: d) {
        if (v == 1) t += v;
      }
    }
    ans = max(ans, t);
  }
  cout << ans << endl;

  return 0;
}