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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
$answer = 0;
$memo = [];
fscanf(STDIN, '%d', $personCnt);
$infoCnts = [];
$infoList = [];
for ($i=0; $i < $personCnt; $i++) { 
    $num = $i;
    fscanf(STDIN, '%d', $infoCnts[$num]);
    for ($j=0; $j < $infoCnts[$num]; $j++) { 
        fscanf(STDIN, '%d %d', $target, $request);
        $infoList[$num][$target-1] = $request;
        if ($request == 1 && !in_array($num.'-'.$target, $memo)) {
            $memo[] = $num.'-'.$target;
        }
    }
}

for ($i=0, $I=2**$personCnt; $i < $I; $i++) { 
    $bits = sprintf('%0'.$personCnt.'b', $i);
    $flag = true;
    foreach ($infoList as $person => $map) {
        if ($bits[$person] == '1') {
            foreach ($map as $key => $value) {
                if ($bits[$key] != $value) {
                    $flag = false;
                }
            }
        }
    }
    if ($flag) {
        $total = 0;
        for ($j=0; $j < $personCnt; $j++) { 
            $total += $bits[$j];
        }
        $answer = max($answer, $total);
    }
}

echo $answer . "\n";