1076 Forwards on Weibo (30分)_assuming that only l levels of indirect followers -优快云博客

本文链接：https://blog.youkuaiyun.com/zbchenchanghao/article/details/104290501

本文介绍了一个基于微博社交网络的算法，用于计算特定用户在限定层级内的最大转发潜力，通过广度优先搜索遍历用户及其间接关注者，评估信息传播范围。

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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 Llevels 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:

Sample Output:

4
5

思路：使用bsf对图进行遍历，保存level，在level小于等于给定的l时，计数未被访问的节点，最后减去自身节点即可。

#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
vector<vector<int> > vec;
bool visit[1002] = {false};
int n, l;
int cnt;
void bfs(int start, int level){
    queue<int> que;
    que.push(start);
    visit[start] = true;
    while (!que.empty())
    {
        
        int z = que.size();
        
        while (z--)
        {
           
            
            for(int i = 0; i < vec[que.front()].size(); i++){

                if(!visit[vec[que.front()][i]] && level <= l){
                    visit[vec[que.front()][i]] = true;
                    que.push(vec[que.front()][i]);
    
                    cnt++;
                }
                
            }
            
            que.pop();
            
        }
        
        level++;
   
        
    }
    


}
int main(){

    
    cin >> n >> l;
    
    vec.resize(n + 1);
    for(int i = 1; i <= n; i++){

        int m;
        scanf("%d", &m);
        for(int j = 0; j < m; j++){
            int temp;
            scanf("%d", &temp);
            vec[temp].push_back(i);

        }
    }
    int k;
    cin >> k;
    for(int i = 0; i < k; i++){
        int id;
        cnt = 1;
        memset(visit, false, sizeof(visit));
        // fill(visit, visit + 1002, false);
        scanf("%d", &id);
        
        bfs(id, 1);
        printf("%d\n", cnt - 1);
    }
   
    return 0;
}