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 Go (1.6) (judge verdict: Accepted)

package main

import "fmt"

func main() {
	var N int
	fmt.Scan(&N)
	shogen := make([][]int, N)
	for i := 0; i < N; i++ {
		shogen[i] = make([]int, N)
		for j := 0; j < N; j++ {
			shogen[i][j] = -1
		}
		var n int
		fmt.Scan(&n)
		for j := 0; j < n; j++ {
			var x, y int
			fmt.Scan(&x, &y)
			x--
			shogen[i][x] = y // 人iからみて人jはy(1:正直,0:不正直)
		}
	}
	maxhonests := 0
	for bits := 0; bits < (1 << uint(N)); bits++ { // 正直な人を決め打って全探索する
		ok := true               // この正直者達の組み合わせで矛盾は生じないか？
		for i := 0; i < N; i++ { // bitsに含まれる正直者の証言をチェックする
			if (bits & (1 << uint(i))) != 0 { // 人iは正直である
				for j := 0; j < N; j++ { // 人iの全ての証言を調べる
					if (shogen[i][j] == 1 && !isHonest(bits, j)) ||
						(shogen[i][j] == 0 && isHonest(bits, j)) {
						// 人iは人jを正直と証言しているが正直者リストに含まれていない、あるいは
						// 人iは人jを不正直と証言しているが、正直者リストに含まれている。
						ok = false // このbitsは矛盾する
					}
				}
			}
		}
		if ok { // 矛盾が生じない場合の正直者の人数を数える
			maxhonests = max(maxhonests, bcount(bits))
		}
	}
	fmt.Println(maxhonests)
}

func isHonest(honests, j int) bool {
	if (honests & (1 << uint(j))) != 0 {
		return true
	}
	return false
}

func bcount(b int) int {
	r := 0
	for b > 0 {
		if (b & 1) != 0 {
			r++
		}
		b /= 2
	}
	return r
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}