CCF python 201709-2 公共钥匙盒

基本思路：
将所有的时间节点（借钥匙时间，还钥匙时间）组织到一个列表里【钥匙号，时间，借or还】
并将之按时间的升序、还优先借、钥匙号的升序排列

那么所要做的就是遍历这个列表：
····如果是借就
········将要钥匙盒对应的位置置零
····如果是还就
········遍历钥匙盒
············找到第一个零，将之变成钥匙号

# 201709-2 	公共钥匙盒
N, K = map(int, input().split())

keybox = []
com = []
all = []
for i in range(N):
    keybox.append(i + 1)
for i in range(K):
    com.append(list(map(int, input().split())))
    all.append([com[i][0], com[i][1], -1])
    all.append([com[i][0], com[i][1] + com[i][2], 1])

all.sort(key=lambda x: (x[1],-x[2],x[0]))

for i in range(len(all)):
    if all[i][2] == -1:  # borrow
        keybox[keybox.index(all[i][0])] = 0
    elif all[i][2] == 1:  # return
        for j in range(len(keybox)):
            if keybox[j] == 0:
                keybox[j] = all[i][0]
                break
for i in range(len(keybox)):
    print(keybox[i], end=' ')

标签：201709,python,len,range,钥匙,keybox,CCF,com,append
来源： https://blog.csdn.net/Ln_Jy/article/details/110562747