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 honest(n int, i uint) int {
	return (n >> i) & 1
}

func honestsNum(n int) int {
	ret := 0
	for i := 0; n > 0; i++ {
		if n&1 == 1 {
			ret++
		}
		n >>= 1
	}
	return ret
}

func max(nums ...int) int {
	ret := nums[0]
	for _, v := range nums {
		if v > ret {
			ret = v
		}
	}
	return ret
}

func main() {
	var n uint
	fmt.Scan(&n)
	a := make([]int, n)
	x := make([][]int, n)
	y := make([][]int, n)
	for i := uint(0); i < n; i++ {
		fmt.Scan(&a[i])
		x[i] = make([]int, a[i])
		y[i] = make([]int, a[i])
		for j := 0; j < a[i]; j++ {
			fmt.Scan(&x[i][j], &y[i][j])
			x[i][j]--
		}
	}

	// bit全探索
	iMax := 1 << n
	ans := 0 // i = 0のとき
	for i := 1; i < iMax; i++ {
		ok := true
		for j := uint(0); j < n; j++ {
			if honest(i, j) == 0 {
				continue
			}
			for k := 0; k < a[j]; k++ {
				if honest(i, uint(x[j][k])) != y[j][k] {
					ok = false
					break
				}
			}
			if !ok {
				break
			}
		}
		if ok {
			ans = max(ans, honestsNum(i))
		}
	}

	fmt.Println(ans)
}