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:7 3 3 2 3 4 0 2 5 6 2 3 1 2 3 4 1 4 1 5 2 2 6Sample Output:
4 5
#include <iostream>
#include <vector>
#include <iterator>
#include <queue>
#include <cstring>
#include <utility>
#define MAX 1005
using namespace std;
struct user{
vector<int> followed;
};
user graph[MAX];
bool visit[MAX];
int N, L;
void init()
{
int num, temp;
for(int i=1; i<=N; i++)
{
cin >> num;
for(int j=0; j<num; j++)
{
cin >> temp;
graph[temp].followed.push_back(i);
}
}
}
void BFS(int x)
{
int counter = 0, level = 0, curlevel = 0;
queue<pair<int, int> > myque;
vector<int>::iterator it;
visit[x] = true;
myque.push(make_pair(x, level));
while(!myque.empty())
{
pair<int, int> temp = myque.front();
myque.pop();
if(curlevel < temp.second)
{
curlevel = temp.second;
}
if(curlevel+1 > L)
break;
for(it=graph[temp.first].followed.begin(); it!=graph[temp.first].followed.end(); it++)
{
if(!visit[*it])
{
visit[*it] = true;
myque.push(make_pair(*it, curlevel+1));
counter++;
}
}
}
cout << counter << endl;
}
int main()
{
int n, temp;
cin >> N >> L;
init();
cin >> n;
while(n--)
{
memset(visit, false, sizeof(bool)*(N+1));
cin >> temp;
BFS(temp);
}
return 0;
}
本文介绍了一个模拟微博用户间转发行为的算法。通过构建社交网络图,使用广度优先搜索(BFS)来计算特定用户发布的内容可能达到的最大转发次数,考虑限定层级内的间接关注者。
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