Python|外卖店优先级

N, M, T = map(int, input().split())

lis = [0 for i in range(N)]   #创建一个基础优先级为0的列表

lis_a, lis_b = [], []

dic = {}

for i in range(M):

    a, b = map(int, input().split())

    lis_a.append(a)

    lis_b.append(b)

for i in range(1, T+1):     #按时间点进行判断

    lis1 = []               #每次遍历的时间点若有订单就加入

    for j in range(M):

        if lis_a[j] == i:

            lis1.append(lis_b[j])

    for k in lis1:         #在列表中，即有订单，+2

        lis[k-1] += 2

    for p in range(1, N+1):  #不在列表，-1

        if p not in lis1 and lis[p-1] > 0:   #最低到0

            lis[p-1] -= 1

    for q in range(N):

        if lis[q] >= 3:   #当优先级>3便加入或者更改

            dic[q+1] = lis[q]

for k in list(dic):   #优先级<=3，便删除

    if int(dic[k]) <= 3:

        dic.pop(k)

print(len(dic))