1076 Forwards on Weibo （30 分)

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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 3000 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一个微博的关注关系网，一个用户发微博时，他的关注者都会转发，关注者的关注者继续转发，求l层内的转发数

分析

虽然30分的题，但是比较简单，用bfs即可。我这里用的是邻接表的结构。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;

vector<int> users[1010];
int see[1010];
int n,l,mi,tmp,k;
deque<int> q;
void cal(int root,int layer){
    q.push_back(root);
    int cnt=0;
    fill(see,see+1010,0);
    see[root]=1;
    while(layer--){
        int qsize=q.size();
        while(qsize--) {
            int cur = q.front();
            q.pop_front();
            for (int u:users[cur]) {
                if (!see[u]) {
                    see[u] = 1;
                    cnt++;
                    q.push_back(u);
                }
            }
        }
    }
    cout<<cnt<<endl;
}
int main() {
    cin>>n>>l;
    for(int i=1;i<=n;i++){
        cin>>mi;
        for(int j=0;j<mi;j++){
            cin>>tmp;
            users[tmp].push_back(i);
        }
    }
    cin>>k;
    for(int i=0;i<k;i++){
        cin>>tmp;
        q.clear();
        cal(tmp,l);
    }
    return 0;
}