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)

/*
 *  Author : Jun_SH
 *  Solution for ABC 147 C
 */
#include<bits/stdc++.h>
#define int long long
using namespace std;
inline int read(){
    int f=1,x=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return f*x;
}
int g[20];
map<int,int>mp;
queue<int>q;
int vis[20]={};
main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin>>n;
    int m;
    int ans=0;
    int flag=0;
    for(int i=1;i<=n;++i){
        if(vis[i])continue;
        g[i]=1;
        mp[0]=0;
        mp[1]=1;
        q.push(i);
        vis[i]=1;
        while(!q.empty()){
            int now=q.front();
            q.pop();
            cin>>m;
            for(int j=1;j<=m;++j){
                int pl,bo;
                cin>>pl>>bo;
                if(vis[pl]){
                    if(g[pl]==g[now])flag=1;
                    continue;
                }
                if(bo==1){
                    g[pl]=g[now];
                }
                else{
                    g[pl]=g[now]^1;
                }
                q.push(pl);
                vis[pl]=true;
            }
        }
        ans+=max(mp[0],mp[1]);
    }
    if(flag)cout<<(0)<<endl;
    else cout<<ans<<endl;
}