PAT A1076 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:

Sample Output:

4

5

 #include <stdio.h>

 #include <algorithm>

 #include <set>

 //#include <string.h>

 #include <vector>

 //#include <math.h>

 #include <queue>

 #include <iostream>

 #include <string>

 using namespace std;

 const int maxn = ;

 const int inf = ;

 int n,k,m,l;

 bool vis[maxn];

 struct node{

     int id,lvl;

 };

 vector<node> adj[maxn];

 int bfs(int root){

     fill(vis,vis+maxn,false);

     int num=;

     queue<node> q;

     node st;

     st.id = root;

     st.lvl = ;

     q.push(st);

     vis[root]=true;

     while(!q.empty()){

         node now = q.front();

         q.pop();

         int u=now.id;

         for(int i=;i<adj[u].size();i++){

             node next = adj[u][i];

             next.lvl = now.lvl+;

             if(vis[adj[u][i].id]==false && next.lvl<=l){

                 q.push(next);

                 vis[next.id]=true;

                 num++;

             }

         }

     }

     return num;

 }

 int main(){

     scanf("%d %d",&n,&l);

     for(int i=;i<=n;i++){

         scanf("%d",&k);

         node user;

         user.id = i;

         for(int j=;j<k;j++){

             scanf("%d",&m);

             adj[m].push_back(user);

         }

     }

     scanf("%d",&k);

     for(int j=;j<k;j++){

         scanf("%d",&m);

         int res=bfs(m);

         printf("%d\n",res);

     }

 }

注意点：图的多次bfs，要注意初始化vis，和dfs不同的是bfs的vis是入队一次就要设为true，而不是访问了才设为true，否则会多次访问到。记录层数需要结构体，和A1106 Lowest Price In Supply Chain 不同，1106需要两个队列来完成一层层保存，因为他是二叉树，而图会存在环，两个队列也可以，不过这题没必要。