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

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers:

M[i] user_list[i]

where M[i] ( $\leq) is the total number of people that user[i] follows; and user_list[i] is a list of the M[i] users that followed by user[i] . It is guaranteed that no one can follow oneself. All the numbers are separated by a space.$

Then finally a positive

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

Sample Input:

Sample Output:

4
5

题不难，但是题意很难懂
下面是对输入的翻译：

每个输入文件包含一个测试用例。对于每种情况，第一行包含2个正整数：N（<= 1000），用户数；L（<= 6），所计数的间接关注者的级别数。因此，假设所有用户的编号都从1到N。然后是N行，每行的格式为：

第一个数表示所关注人的个数M，后面紧跟M个ID，表示所关注人的ID。所有数字都用空格分隔。

然后，最后给出一个正数K，后跟K个用户ID进行查询。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
#define  inf  0x3fffffff
typedef struct Node{
    int id;
    int level;
}node;
vector<vector<int> > v;
int n,L;
int bfs(node tnode){
    int cnt=0;
    bool vis[maxn];
    fill(vis,vis+maxn,false);
    queue<node> q;
    q.push(tnode);
    vis[tnode.id]=true;
    while(!q.empty()){
        node t=q.front();
        q.pop();
        for(int i=0;i<v[t.id].size();i++){
            int next=v[t.id][i];
            if(vis[next]==false&&t.level<L){
                node tnext={next,t.level+1};
                q.push(tnext);
                vis[next]=true;
                cnt++;
            }
        }
    }
    return cnt;
}

int main(){
    scanf("%d %d",&n,&L);
    v.resize(n+1);
    for(int i=1;i<=n;i++){
        int k;
        scanf("%d",&k);
        for(int j=0;j<k;j++){
            int mm;
            scanf("%d",&mm);
            v[mm].push_back(i);
//            v[i].push_back(mm);
        }
    }
    int m,t;
    scanf("%d",&m);
    for(int i=0;i<m;i++){
        scanf("%d",&t);
        node tnode={t,0};
        printf("%d
",bfs(tnode));
    }
    return 0;
}