hihocoder，#1497 : Queen Attack。python

 

N = int(raw_input())
data = []
x_ = []
y_ = []
for i in range(N):
    dataline = raw_input().strip('\n').split(" ")
    dataline = map(int, dataline)
    x, y = dataline[0], dataline[1]
    data.append([x, y])
    x_.append(x)
    y_.append(y)

add = []  # x+y
minus = []  # x-y
for d in data:
    add.append(d[0] + d[1])
    minus.append(d[0] - d[1])

dictx_ = {}
dicty_ = {}
dictadd_ = {}
dictminus_ = {}

for d in x_:
    if d not in dictx_:
        dictx_[d] = 1
    else:
        dictx_[d] += 1

for d in y_:
    if d not in dicty_:
        dicty_[d] = 1
    else:
        dicty_[d] += 1

for d in add:
    if d not in dictadd_:
        dictadd_[d] = 1
    else:
        dictadd_[d] += 1

for d in minus:
    if d not in dictminus_:
        dictminus_[d] = 1
    else:
        dictminus_[d] += 1

ans = 0

for d in dictx_:
    if dictx_[d] != 1:
        ans += dictx_[d] * (dictx_[d] - 1) / 2

for d in dicty_:
    if dicty_[d] != 1:
        ans += dicty_[d] * (dicty_[d] - 1) / 2

for d in dictadd_:
    if dictadd_[d] != 1:
        ans += dictadd_[d] * (dictadd_[d] - 1) / 2

for d in dictminus_:
    if dictminus_[d] != 1:
        ans += dictminus_[d] * (dictminus_[d] - 1) / 2

print ans

思路：

1、对N个点的横坐标、纵坐标、横纵坐标之和、横纵坐标之差分别建立字典，然后统计对键值为2以上的个数。比如横坐标为4的有4个点，那么就有4*(4-1)/2=6个，即C4  2