A1076 Forwards on Weibo（30分）PAT 甲级(Advanced Level) Practice（C++）满分题解【BFS】

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

题意分析：

题目给出的是每个用户关注的人，所以在储存到数组时应该转换为每个人的粉丝群体，这样才能直接表示一个人可以带来的转发量，然后从查询用户为起点，广度优先遍历该用户的粉丝列表，然后粉丝列表的粉丝列表，当层数超过L后就不计算了。

代码如下：

#include
using namespace std;
int N, L,qn,quid,maxForwards;
vectorint>> M;//记录用户i的粉丝
vector<bool> isvisit;
int bfs(int s)
{
	queue<int> que;
	que.push(s);
	isvisit[s] = true;
	int level = -1, last = s, tail;
	int count = -1;
	while (que.empty()==false)
	{
		int v = que.front();
		que.pop();
		for (int i = 0; i < M[v].size(); i++)
		{
			if (isvisit[M[v][i]] == false) {
				isvisit[M[v][i]] = true;
				que.push(M[v][i]);
				tail = M[v][i];
			}
		}
		count++;
		if (last == v)
		{
			level++;
			last = tail;
		}
		if (level == L)
			return count;
	}
	return count;
}
int main()
{
	cin >> N >> L;
	M.resize(N + 1);
	isvisit.resize(N + 1);
	int num,uid;
	for (int i = 1; i <= N; i++)
	{
		cin >> num;
		for (int j = 1; j <= num; j++) {
			cin >> uid;
			M[uid].push_back(i);
		}
	}
	cin >> qn;
	for (int i = 0; i < qn; i++) {
		cin >> quid;
		maxForwards = 0;
		fill(isvisit.begin(), isvisit.end(), false);
		/*for (int j = 0; j < M[quid].size(); j++) {
			dfs(M[quid][j], 0, maxForwards);
		}*/
		cout << bfs(quid)<
	}
	return 0;
}

运行结果如下：