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Forwards on Weibo (30)

[编程题]Forwards on Weibo (30)
  • 热度指数:5017 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 64M,其他语言128M
  • 算法知识视频讲解
Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may follow many other users as well. Hence a social network is formed with followers relations. When a user makes a post on Weibo, all his/her followers can view and forward his/her post, which can then be forwarded again by their followers. Now given a social network, you are supposed to calculate the maximum potential amount of forwards for any specific user, assuming that only L levels of indirect followers are counted.

输入描述:
Each input file contains one test case.  For each case, the first line contains 2 positive integers: N (<=1000), the number of users; and L (<=6), the number of levels of indirect followers that are counted.  Hence it is assumed that all the users are numbered from 1 to N.  Then N lines follow, each in the format:
M[i] user_list[i]
where M[i] (<=100) is the total number of people that user[i] follows; and user_list[i] is a list of the M[i] users that are followed by user[i]. It is guaranteed that no one can follow oneself. All the numbers are separated by a space.
Then finally a positive K is given, followed by K UserID's for query.


输出描述:
For each UserID, you are supposed to print in one line the maximum potential amount of forwards this user can triger, assuming that everyone who can view the initial post will forward it once, and that only L levels of indirect followers are counted.
示例1

输入

7 3
3 2 3 4
0
2 5 6
2 3 1
2 3 4
1 4
1 5
2 2 6

输出

4
5
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发表于 2021-09-22 14:36:36
更多PAT甲级题解尽在我的个人博客--acking-you.github.io 题目 OJ平台 题目大意 这题看了描述后,大概就能清楚题目是这么个意思: 输入: 输入N个人,并给出这N个人的关注列表。以及层级限制 L 和 K 次询问。 输出: K次询问,每次询问第 i 个人发博客后,在它树形关 展开全文