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 Julia (0.5.0) (judge verdict: Runtime Error)

#=
C:
- Julia version: 
- Author: abap
- Date: 2019-12-21
=#

function return_all_pettern(x)  #bit全探索
    count_pettern = x ^ 2
    b = []
    for i in 0:count_pettern - 1
        bit = bin(i, x)
        bits = []
        for j in 1:x
            push!(bits, bit[j])
        end
        bits = [bits .== '1']
        push!(b, bits)
    end
    #println(length(b))
    return  b
end


function main()
XY = Dict{Int64, Array{Any,1}}() # xıとyıの格納先
N = parse(Int,readline())
for i in 1:N
    A = parse(Int,readline())
    for j in  1:A
       xy = parse.(split(readline()))
       xy[2] = xy[2] == 1
       push!(XY,i=>xy)       # i = 発言者 xy = 主張
    end
end
petterns = return_all_pettern(length(keys(XY)))  # 全ての部分集合を取得
res_said = []
for i in 1:length(petterns)               # それぞれの部分集合についての結果の格納先
    push!(res_said,[])
end
for jj in 1:length(petterns)        # 特定の格納先に入れるためにインデックスを回す
    pettern = petterns[jj]
    honest_list = collect(1:N)[sum(pettern)]
    for k in honest_list
        push!(res_said[jj],[k,1])      # 発言している人は本物
        push!(res_said[jj], [XY[k][1],Bool(XY[k][2]) ]  ) # 本物の言っている人も本物
    end
end
global kari_ans = []
for i in 1:length(petterns)               # それぞれの仮の答えについての結果の格納先
    push!(kari_ans,[])
end

for kk in 1:length(petterns)        # 特定の格納先に入れるためにインデックスを回す
    res = res_said[kk]
    if res == []
        ans = 0
    else
        for ik in res
            push!(kari_ans[kk],[ik[1],ik[2]])
        end
    end
end
ans_list  = []
for kari in kari_ans
    #println(kari)
    honest_count = 0
    for i in kari  # それぞれの結果について検証する
        #println(i)
        if i[2] == 0
            another_ans = [i[1],1]  # 違う答えがないかチェック
        else
            another_ans = [i[1],0]
            count = 0
            for jj in kari
                if jj == i
                    count += 1
                end
            end
            if count == 1
                honest_count += 1 # 同じ答えが何回もあるときはカウントしない
            end
        end
        if another_ans in kari
            honest_count = 0
            break
        end
    end
    push!(ans_list,honest_count)
end
println(maximum(ans_list))
end

main()