1076. Forwards on Weibo (30)

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.

Input Specification:

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; anduser_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.

Output Specification:

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.

Sample Input:

7 33 2 3 402 5 62 3 12 3 41 41 52 2 6

Sample Output:

45

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指定可以计算l层关注的关系，求出某个人的粉丝（直接的和间接的都计算在内）有多少个。其实就是图的问题。储存每个人的直接粉丝有哪些（注意输入的是每个人关注了哪些人）。然后输入的要查找的人，从这个人出发，用广度优先搜索查找到l层，计算有多少个点，这个就是某个人的粉丝数。

代码：

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>using namespace std;int main(){int n,l;cin>>n>>l;vector<vector<int> >follower(n+1);for(int i=1;i<=n;i++){int m;cin>>m;for(int j=0;j<m;j++){int a;cin>>a;follower[a].push_back(i);}}int k;cin>>k;for(int i=0;i<k;i++){int q;cin>>q;vector<bool>isvis(n+1,false);isvis[q]=true;queue<int>que;que.push(q);int count=0,level=l;while(level>0){int qsize=que.size();for(int j=0;j<qsize;j++){int tmp=que.front();que.pop();for(int t=0;t<follower[tmp].size();t++){if(!isvis[follower[tmp][t]]){que.push(follower[tmp][t]);count++;isvis[follower[tmp][t]]=true;}}}level--;}cout<<count<<endl;}}