在OpenJudge看到一个题目（#4109），题目描述如下：
小明和小红去参加party。会场中总共有n个人，这些人中有的是朋友关系，有的则相互不认识。朋友关系是相互的，即如果A是B的朋友，那么B也是A的朋友。小明和小红想知道其中某两个人有多少个公共的朋友。
输入第一行为一个正整数c，代表测试数据的个数。接下来是c组测试数据。
对于每组测试数据，第一行是三个数字n(2<=n<=100)，m和k，分别表示会场中的人数，已知的朋友关系数目，问题的数目。接下来的m行，每行用两个数字i和j(1<=i,j<=n)表示了一个朋友关系，表示第i个人和第j个人是朋友关系。接下来的k行，每行用两个数字i和j(1<=i,j<=n)表示一个问题，请问第i个人和第j个人有多少公共的朋友。输出对于第i组测试数据，首先输出一行”Case i:”，接下来得k行代表了k个问题，每行输出第i个人和第j个人有多少公共的朋友。
用Python写了段代码，大致实现：
# -*- coding:utf-8 -*- def set_1(i, q):
''' generate a i*i ARRAY for all relationships
if there is a relation set 1 or 0
return i*i ARRAY with 1 or 0
'''
a = [0 for i in range(i*i)]
for j in range(len(q)):
n, m = q[j]
a[(n-1)*i+(m-1)] = 1
a[(m-1)*i+(n-1)] = 1
return a def solve(i, q, r):
''' solve question
i is the number of people
q is the set of questions
r is the set of relationships, the result of function set_1();
'''
result = 0
for j in range(len(r)):
n, m = r[j]
for l in range(i):
if q[(n-1)*i+l] == 1 and q[(m-1)*i+l] == 1:
result += 1
print(result)
result = 0 def main():
d = [ 3, [3,2,2],
[1,3],
[2,3],
[1,2],
[1,3],
[4,3,2],
[1,2],
[2,3],
[1,4],
[2,4],
[1,3],
[5,2,1],
[1,2],
[1,4],
[3,4]
]
for x in d: #Dispaly input
print(x) loc = []
for m in range(1,len(d)): #Get the index of every question
if len(d[m])==3:
loc.append(m) for i in range(len(loc)): #Sovle each question
pNum, qNum, aNum = d[loc[i]] #slice out R and Q in d[]
t = loc[i]+1
R = d[t:t+qNum]
Q = d[t+qNum:t+qNum+aNum] r_1 = set_1(pNum,R) # set 1 for question
print('-------------------\nCase'+str(i+1)+':')
solve(pNum, r_1, Q)
if __name__=='__main__':
main() '''OUTPUT
3
[3, 2, 2]
[1, 3]
[2, 3]
[1, 2]
[1, 3]
[4, 3, 2]
[1, 2]
[2, 3]
[1, 4]
[2, 4]
[1, 3]
[5, 2, 1]
[1, 2]
[1, 4]