python实现CCF 201912-2 《回收站选址》

n = int(input())
loc_list = []
for i in range(n):
    a = input().split(' ')
    a1 = int(a[0])
    a2 = int(a[1])
    loc_list.append([a1,a2])

#检查邻居位置是否存在垃圾
def check_loc(loc,loc_list):
    x = loc[0]
    y = loc[1]
    if [x-1,y] in loc_list and [x,y-1] in loc_list and [x+1,y] in loc_list and [x,y+1] in loc_list:         #逆时针方向判断
        return 1
    else:
        return 0

#检查角落位置是否存在垃圾，并计数
def count_one(loc,loc_list):
    corner_num = 0
    x = loc[0]
    y = loc[1]
    if [x-1,y+1] in loc_list:   #逆时针方向判断
        corner_num +=1
    if [x-1,y-1] in loc_list:
        corner_num +=1
    if [x+1,y-1] in loc_list:
        corner_num +=1
    if [x+1,y+1] in loc_list:
        corner_num +=1
    return corner_num

def count_all(loc_num,count_all_num):
    if loc_num == 0:
        count_all_num[0] +=1
    elif loc_num == 1:
        count_all_num[1] +=1
    elif loc_num == 2:
        count_all_num[2] +=1
    elif loc_num == 3:
        count_all_num[3] +=1
    elif loc_num == 4:
        count_all_num[4] +=1   
    return count_all_num

count_all_num = []
for i in range(5):
    count_all_num.append(0)

for i in range(n):
    loc_status = check_loc(loc_list[i],loc_list)
    if loc_status == 1:
        loc_num = count_one(loc_list[i],loc_list)
        count_all_num = count_all(loc_num,count_all_num)

for i in range(5):
    print(count_all_num[i])

流程图如下