1076. Forwards on Weibo (30)

最新推荐文章于 2020-03-28 18:49:18 发布

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本文介绍了一道关于模拟微博转发过程的算法题目。通过图的限制深度广度优先搜索(BFS)来计算特定用户微博的最大潜在转发数量，考虑到间接关注者的层级限制。

1076. Forwards on Weibo (30)

时间限制

3000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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:

Sample Output:

4
5

这题并不难，图的限制深度的BFS。重要的是邻接表的表示方法，本题

使用 vector a[1005] 来表示图，十分方便。

因为给的信息是用户i follow了谁，而我们要看某个用户的微博可以被谁看，所以用邻接表存的时候反顾来存，并且是单向的

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
vector<int> G[1005];
int vis[1005];
int n,level;
int bfs(int a){
    memset(vis,0,sizeof(vis));
    int first = 0,last = 1,newlevel = 0,cnt = 1;
    //这几个变量特别解释一下，first和last是判断这一层是否输出完，因为第一个传进来的是用户本身不能算，所以newlevel从0开始，first从0开始，
    //last和cnt从1开始是为了过滤去开始的点即用户本身，便于循环，最后只需要减去1即可
    queue<int> q;
    q.push(a);
    vis[a] = 1;
    while(!q.empty()){
        int s = q.front();
        q.pop();
        first++;//每次取出一个++，包含了用户本身
        for(int i = 0; i < G[s].size(); i++){
            if(!vis[G[s][i]]){
                vis[G[s][i]] = 1;
                cnt++;//计算总数，同时得到了加上这一层后的总数
                q.push(G[s][i]);
            }
        }
        if(first == last){
            last = cnt;
            newlevel++;
        }
        if(newlevel == level){
            return cnt-1;
        }
    }
    return cnt-1;
}
int main(){
    cin >> n >> level;
    for(int i = 1; i <= n; i++){
        int m;
        cin >> m;
        while(m--){
            int t;
            cin >> t;
            G[t].push_back(i);
        }
    }
    int k;
    cin >> k;
    while(k--){
        int user;
        cin >> user;
        cout << bfs(user) << endl;
    }
    return 0;
}