1076 Forwards on Weibo (30分) BFS

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针对Weibo社交网络，计算特定用户在限定转发层级内的最大转发量。利用BFS算法遍历用户关注网络，统计直接与间接关注者的转发潜力。

文章目录

问题
解决方法

问题

1076 Forwards on Weibo (30分)

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; and user_list[i] is a list of the M[i] users that 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 trigger, 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 3
3 2 3 4
0
2 5 6
2 3 1
2 3 4
1 4
1 5
2 2 6

Sample Output:

4
5

解决方法

题目大意

给出每个⽤户关注的⼈的id，和转发最多的层数，求⼀个id发了条微博最多会有多少个⼈转发

解题思路

标准的BFS题目，套用BFS的模板即可。

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
const int maxn = 1010;
struct node
{
	int v;
	int layer;
}user;
vector<node>adj[maxn];
bool inq[maxn] = { false };
int n, layernum;
int BFS(int v)
{
	int numforward = 0;
	queue<node>q;
	node start;
	inq[v] = true;
	start.v = v;
	start.layer = 0;
	q.push(start);
	while(!q.empty())
	{
		node topnode = q.front();
		q.pop();
		int u = topnode.v;
		for (int i = 0; i < adj[u].size(); i++)
		{
			node next = adj[u][i];
			next.layer = topnode.layer + 1;
			if (inq[next.v] == false && next.layer <= layernum)
			{
				q.push(next);
				inq[next.v] = true;
				numforward++;
			}
		}
	}
	return numforward;
}
int main()
{
	
	scanf("%d %d", &n, &layernum);
	for (int i = 1; i <= n; i++)
	{
		int k, t;
		scanf("%d", &k);
		user.v = i;
		for (int j = 0; j < k; j++)
		{
			scanf("%d", &t);
			adj[t].push_back(user);
		}
	}
	int query, id;
	scanf("%d", &query);
	for (int i = 0; i < query; i++)
	{
		fill(inq, inq + maxn, false);
		scanf("%d", &id);
		cout << BFS(id) << endl;
	}
	return 0;
}