个人学习记录，代码难免不尽人意。
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
#include<cstdio>
#include<queue>
using namespace std;
const int maxn=1020;
int G[maxn][maxn]={0};
int N,L;
int BFS(int a,bool vis[]){queue<int> q;q.push(a);vis[a]=true;int layer[N]={0};layer[a]=0;int num=0;while(!q.empty()){int u=q.front();q.pop();if(layer[u]<=L){if(u!=a) num++;for(int k=1;k<=N;k++){if(G[u][k]!=0&&vis[k]==false){q.push(k);vis[k]=true;layer[k]=layer[u]+1; }} } }return num;
}int main(){scanf("%d%d",&N,&L);for(int i=1;i<=N;i++){int n;scanf("%d",&n);for(int j=0;j<n;j++){int a;scanf("%d",&a);G[a][i]=1;}}int m;scanf("%d",&m);for(int i=0;i<m;i++){bool vis[maxn]={false};int a;scanf("%d",&a);printf("%d\n",BFS(a,vis));}
}
本题是利用BFS算法来解决，题干中多加了一些限制条件：设置了L层，输出节点个数等等。本题需要注意的是vis这个bool数组需要重复利用，所以最好不要将其设置为全局变量（设置全局变量也可以，就是要重新赋值）。
除此之外，题目给的输入是给了关注者列表，但是我们遍历的时候是从给出节点的追随者（粉丝）来遍历的，所以读入的时候赋值时数组下标是反过来的，需要注意。
