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)


/*
    Bismillahir Rahmanir Raheem
    Think Twice Code Once
    author : reaziii
*/
#include<bits/stdc++.h>
#define sc(x)        scanf("%lld",&x)
#define pcase(x)     printf("Case %d: ",x++)
#define pn           printf("\n")
#define fri(f)       for(int i=0;i<f;i++)
#define frj(f)       for(int j=0;j<f;j++)
#define reset(x,e)   memset(x,e,sizeof(x))
#define all(x)       x.begin(),x.end()
#define input        freopen("input.txt","r",stdin);
#define output       freopen("output.txt","w",stdout)
#define infi         INT_MAX
#define m_v          3000000
#define pii          pair<int,int>
#define pch          pair<char,char>
#define pll          pair<ll,ll>
#define mgraph       map<int,set<int> >
#define mp           make_pair
#define pb           push_back
#define clr(x)       memset(x,0,sizeof(x))
#define frx(x,y)     for(int x=0;x<y;x++)
#define deb(x)       cout<<x
#define ideb(i,x)    cout<<i<<" "<<x
#define F            first
#define S            second
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
bool checkbitt(ll num,int pos);
ll setbitt(ll num,ll pos);
ll resetbitt(ll num,int pos);
 //.................___Start from here___.................//
//.................._____________________................//
 
 
ll bitcount(ll n){
    int ans = 0;
    for(int i=0;i<32;i++){
        ans+=(n>>i)&1;
    }

    return ans;
}
 
 
 
 
void solve(){
    int n;
    cin>>n;
    int mask = 1<<n;
    mask--;
 
    int a[n+1],x[n+1][n+1],y[n+1][n+1];
 
    for(int u = 1 ; u <= n ; u++ ){
        cin>>a[u];
        for(int v = 1;v<=a[u];v++){
            cin>>x[u][v]>>y[u][v];
        }
    }
 
    ll ans = 0;
 
    for(int i=1;i<=mask;i++){
        bool f =true;
 
        for(int j=1;j<=n;j++){
            if(checkbitt(i,j-1)){
                for(int k=1;k<=a[j];k++){
                    if(y[j][k]^checkbitt(i,(x[j][k]-1))) f = false;
                }
            }
        }
 
        if(f) ans = max(ans,bitcount(i));
    }
 
    cout<<ans<<endl;
 
}
 
 
int main(){
    // input;
    // output;
 
    int cs=1;
    int cn=1;
    // cin>>cs;
    while(cs--){
        solve();
    }
}
 
 
 
 
 
 
bool checkbitt(ll num,int pos){
    return (num>>pos) & 1;
}
 
ll setbitt(ll num,ll pos){
    return (1<<pos) | num;
}
 
ll resetbitt(ll num,int pos){
    if(!checkbitt(num,pos)) {
        return num;
    }
    else{
        return (1<<pos)^num;
    }
}