# BOJ - 15089 - Is-A? Has-A? Who Knowz-A?

Updated:

``````import sys

def solution():
data_ = {}

is_a = [[0 for _ in range(500)] for _ in range(500)]
has_a = [[0 for _ in range(500)] for _ in range(500)]

index = 0
for _ in range(N):
C1, R, C2 = map(str, sys.stdin.readline().split())

if C1 in data_: pass
else:
data_[C1] = index
index += 1

if C2 in data_: pass
else:
data_[C2] = index
index += 1

if R == "is-a":
is_a[data_[C1]][data_[C2]] = 1
elif R == "has-a":
has_a[data_[C1]][data_[C2]] = 1

for k in range(len(data_)):
for i in range(len(data_)):
for j in range(len(data_)):
is_a[i][j] = is_a[i][j] | (is_a[i][k] & is_a[k][j])
has_a[i][j] = has_a[i][j] | (has_a[i][k] & has_a[k][j])
has_a[i][j] = has_a[i][j] | (is_a[i][k] & has_a[k][j])
has_a[i][j] = has_a[i][j] | (has_a[i][k] & is_a[k][j])

for i in range(1, M+1):
C1, R, C2 = map(str, sys.stdin.readline().split())

if (C1 == C2) and R == "is-a":
print("Query {}: true".format(i))
continue

if R == "is-a":
if is_a[data_[C1]][data_[C2]]:
print("Query {}: true".format(i))
else:
print("Query {}: false".format(i))
elif R == "has-a":
if has_a[data_[C1]][data_[C2]]:
print("Query {}: true".format(i))
else:
print("Query {}: false".format(i))
solution()

``````

https://www.acmicpc.net/problem/15089

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