Problem: 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?

Submission attempts (in order):
def checker(bbit,listN,N):
    i = 0
    j = 0
    ng = False
    #print(N)
    #print(listN)
    while N > i :
        j=0
        checkFig = 3
        ibit = 1 << i
        while N > j :
            cbit = 1 << j
            if bbit & cbit == 0 :
                #print("A:{} {}".format(j,i))
                pass
            elif listN[j][i] == 3 :
                #print("B:{} {}".format(j,i))
                pass
            elif (listN[j][i] == 0) and (bbit & ibit != 0):
                ng = True
                break
            elif checkFig == 3 :
                #print("C:{} {}".format(j,i))
                checkFig = listN[j][i]
            elif checkFig != listN[j][i] :
                ng = True
                #print("D:{} {}".format(j,i))
                break
            j += 1
        if ng :
            #print(ng)
            break
        else :
            i +=1
    return (not ng)


def main():
    N = int(input())
    listA = [[3 for j in range(N)] for k in range(N)]
    n = 0
    while N > n :
        a = int(input())
        i = 0
        while a > i:
           x,y  = map(int,input().split())
           listA[n][x-1] = y
           i += 1
        n += 1
        
    honestBit = pow(2, N)-1
    honest = N
    while honest > 0:
        ok = checker(honestBit,listA,N)
        if ok :
            break
        honestBit -= 1
        honest -= 1

    print(honest)

main()


def checker(bbit,listN,N):
    i = 0
    j = 0
    ng = False
    #print(N)
    #print(listN)
    
    while N > i :
        j=0
        checkFig = 3
        ibit = 1 << i
        while N > j :
            cbit = 1 << j
            if bbit & cbit == 0 :
                #print("A:{} {}".format(j,i))
                pass
            elif listN[j][i] == 3 :
                #print("B:{} {}".format(j,i))
                pass
            elif (listN[j][i] == 0) and (bbit & ibit != 0):
                ng = True
                break
            elif checkFig == 3 :
                #print("C:{} {}".format(j,i))
                checkFig = listN[j][i]
            elif checkFig != listN[j][i] :
                ng = True
                #print("D:{} {}".format(j,i))
                break
            j += 1
        if ng :
            #print(ng)
            break
        else :
            i +=1
    if N == i:
        ng =True
    return (not ng)


def main():
    N = int(input())
    listA = [[3 for j in range(N)] for k in range(N)]
    n = 0
    while N > n :
        a = int(input())
        i = 0
        while a > i:
           x,y  = map(int,input().split())
           listA[n][x-1] = y
           i += 1
        n += 1
        
    honestBit = pow(2, N)-1
    honest = 0
    
    while honestBit > 0:
        ok = checker(honestBit,listA,N)
        if ok :
            okbit = honestBit
            oknum = 0
            while okbit > 0 :
                oknum += (okbit % 2) 
                okbit = okbit >> 2        
            if oknum > honest :
                honest = oknum 
        honestBit -= 1
    print(honest)
    

main()


def checker(bbit,listN,N):
    i = 0
    j = 0
    ng = False
    #print(N)
    #print(listN)
    
    while N > i :
        j=0
        checkFig = 3
        ibit = 1 << i
        while N > j :
            cbit = 1 << j
            if bbit & cbit == 0 :
                #print("A:{} {}".format(j,i))
                pass
            elif listN[j][i] == 3 :
                #print("B:{} {}".format(j,i))
                pass
            elif (listN[j][i] == 0) and (bbit & ibit != 0):
                ng = True
                break
            elif (listN[j][i] == 1) and (bbit & ibit == 0):
                ng = True
                break
            elif checkFig == 3 :
                #print("C:{} {}".format(j,i))
                checkFig = listN[j][i]
            elif checkFig != listN[j][i] :
                ng = True
                #print("D:{} {}".format(j,i))
                break
            j += 1
        if ng :
            #print(ng)
            break
        else :
            i +=1
    return (not ng)


def main():
    N = int(input())
    listA = [[3 for j in range(N)] for k in range(N)]
    n = 0
    while N > n :
        a = int(input())
        i = 0
        while a > i:
           x,y  = map(int,input().split())
           listA[n][x-1] = y
           i += 1
        n += 1
        
    honestBit = pow(2, N)-1
    honest = 0
    
    while honestBit > 0:
        ok = checker(honestBit,listA,N)
        if ok :
            okbit = honestBit
            oknum = 0
            while okbit > 0 :
                oknum += (okbit % 2) 
                okbit = okbit >> 1
            if oknum > honest :
                honest = oknum 
        honestBit -= 1
    print(honest)
    

main()

Verdicts: Wrong Answer

Wrong Answer

Accepted