1076 Forwards on Weibo

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 (≤), the number of users; and L (≤), 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] (≤) 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：

　　给出一个社交网络，在规定的层级内，一个博主发了博文之后，共有多少人能够看到这篇文章？假设每一个follower看到发表的文章之后都会转发。

思路：

　　这道题就是一道有向图的问题，样例中：M[i] user_list[i]代表的意思应该是，user_list中的人发表了文章，i能够看到并进行转发。然后用邻接矩阵表示有向图，用层序遍历的方法来遍历满足要求的结点。注意遍历的过程中不要重复出现。

Code：

#include <bits/stdc++.h>

using namespace std;

vector<int> grap[1005];

int helper(int query, int level) {
    queue<int> que;
    que.push(query);
    que.push(-1);
    int count = 0;
    vector<bool> visited(1005, false);
    visited[query] = true;
    while (!que.empty() && level >= 0) {
        int temp = que.front();
        que.pop();
        if (temp != -1) count++;
        if (temp == -1) {
            level--;
            if (que.empty()) break;
            que.push(-1);
        } else {
            for (int i = 0; i < grap[temp].size(); ++i) {
                if (!visited[grap[temp][i]]) {
                    visited[grap[temp][i]] = true;
                    que.push(grap[temp][i]);
                }
            }
        }
    }
    return count - 1;
}

int main() {
    int n, l, k, t;
    cin >> n >> l;
    for (int i = 1; i <= n; ++i) {
        cin >> k;
        for (int j = 0; j < k; ++j) {
            cin >> t;
            grap[t].push_back(i);
        }
    }
    cin >> k;
    for (int i = 0; i < k; ++i) {
        cin >> t;
        cout << helper(t, l) << endl;
    }

    return 0;
}