CCF 201912-2 回收站选址 python 满分

题目叙述

问题描述：略

输入格式：略

输出格式：略

样例

样例输入
7
1 2
2 1
0 0
1 1
1 0
2 0
0 1

样例输出
0
0
1
0
0

样例输入
2
0 0
-100000 10

样例输出
0
0
0
0
0

样例输入
11
9 10
10 10
11 10
12 10
13 10
11 9
11 8
12 9
10 9
10 11
12 11

样例输出
0
2
1
0
0

满分证明

解题思路

1. 思路应该特别简单，只是我想难了；
2. 直接存在list中，用in去判断就好了；
3. 另一种就是用数组模拟整个坐标系（坐标系小还可以尝试），但这样内存会爆，下面附代码，你可以试试，(#^.#)！

满分代码

满分代码一

ss = []
n = int(input())
fc = [0] * 5


def sxzy(mzb):
    tx = mzb[0]
    ty = mzb[1]
    if ([tx, ty + 1] in ss) & ([tx, ty - 1] in ss) & ([tx + 1, ty] in ss) & ([tx - 1, ty] in ss):
        return True
    else:
        return False


def js(pzb):
    te = 0
    tx = pzb[0]
    ty = pzb[1]
    # 右上
    if [tx + 1, ty + 1] in ss:
        te += 1
    # 左下
    if [tx - 1, ty - 1] in ss:
        te += 1
    # 右下
    if [tx + 1, ty - 1] in ss:
        te += 1
    # 左上
    if [tx - 1, ty + 1] in ss:
        te += 1
    return te


for _ in range(n):
    ss.append(list(map(int, input().split())))
for i in range(n):
    if sxzy(ss[i]):
        if js(ss[i]) == 0:
            fc[0] += 1
        elif js(ss[i]) == 1:
            fc[1] += 1
        elif js(ss[i]) == 2:
            fc[2] += 1
        elif js(ss[i]) == 3:
            fc[3] += 1
        elif js(ss[i]) == 4:
            fc[4] += 1
for i in range(5):
    print(fc[i])

满分代码二

n = int(input())
zb=[]
count=[0,0,0,0,0]

for _ in range(n):
    x,y=map(int,input().split())
    zb.append([x,y])
for zz in zb:
    ct=0
    if ([zz[0],zz[1]+1] in zb ) &([zz[0],zz[1]-1] in zb )&([zz[0]-1,zz[1]] in zb )&([zz[0]+1,zz[1]] in zb ):
        if [zz[0]-1,zz[1]+1] in zb:
            ct=ct+1
        if [zz[0]-1,zz[1]-1] in zb:
            ct=ct+1                                                                                      
        if [zz[0]+1,zz[1]-1] in zb:
            ct=ct+1 
        if [zz[0]+1,zz[1]+1] in zb:
            ct=ct+1
        count[ct]=count[ct]+1
for cc in count:
    print(cc)  

思路正确，但爆内存

zb = ["0" * (2 * 10 ** 9 + 3) for _ in range(2 * 10 ** 9 + 3)]
ss = []
n = int(input())
fc = [0] * 5


def sxzy(mzb):
    tx = mzb[0]
    ty = mzb[1]
    if zb[tx][ty + 1] == zb[tx][ty - 1] == zb[tx + 1][ty] == zb[tx - 1][ty] == "1":
        return True
    else:
        return False


def js(pzb):
    te = 0
    tx = pzb[0]
    ty = pzb[1]
    if zb[tx - 1][ty + 1] == "1":
        te += 1
    if zb[tx + 1][ty - 1] == "1":
        te += 1
    if zb[tx + 1][ty + 1] == "1":
        te += 1
    if zb[tx - 1][ty - 1] == "1":
        te += 1
    return te


for _ in range(n):
    jx, jy = map(int, input().split())
    x = 10 ** 9 + 1 - jy
    y = 10 ** 9 + 1 + jx
    zb[x] = zb[x][:y] + "1" + zb[x][y + 1:]
    ss.append([x, y])
for i in range(n):
    if sxzy(ss[i]):
        if js(ss[i]) == 0:
            fc[0] += 1
        elif js(ss[i]) == 1:
            fc[1] += 1
        elif js(ss[i]) == 2:
            fc[2] += 1
        elif js(ss[i]) == 3:
            fc[3] += 1
        elif js(ss[i]) == 4:
            fc[4] += 1
for i in range(5):
    print(fc[i])

需者自取传送门(∩ᄑ_ᄑ)⊃━☆【CCF 2013-2021】本博主整理历年至少前两题 python 满分代码目录