PAT (Advanced Level) 1076. Forwards on Weibo (30) 微博转发 BFS

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本文介绍了一个模拟微博用户间转发行为的算法。通过构建社交网络模型，使用BFS算法计算特定用户在限定层级内的最大转发潜力。文章提供了完整的C++实现代码。

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

记录每个结点的粉丝，进行L次BFS。用哈希表记录访问过的结点防止重复计算。

/*2015.7.29cyq*/
#include <iostream>
#include <vector>
#include <queue>
#include <fstream>
using namespace std;

//ifstream fin("case1.txt");
//#define cin fin

int main(){
	int N,L;
	cin>>N>>L;
	vector<vector<int> > tree(N+1);
	int count,x;
	for(int i=1;i<=N;i++){//记录i关注的人
		cin>>count;
		for(int j=1;j<=count;j++){
			cin>>x;
			tree[x].push_back(i);//i是x的粉丝
		}
	}
	int M;
	cin>>M;
	int result;
	while(M--){
		result=0;
		cin>>x;
		vector<bool> visited(N+1,false);
		queue<int> cur,next;
		cur.push(x);
		visited[x]=true;
		for(int i=0;i<L;i++){//前L层粉丝
			while(!cur.empty()){
				int tmp=cur.front();
				cur.pop();
				for(auto it=tree[tmp].begin();it!=tree[tmp].end();it++){
					if(!visited[*it]){
						visited[*it]=true;
						next.push(*it);
						result++;
					}
				}
			}
			swap(cur,next);
		}
		cout<<result<<endl;
	}
	return 0;
}