一、题目描述
原题链接
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:
Output Specification:
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
二、解题思路
给定起点，图的层序遍历题。注意题目这里给出的数据是一个用户follow的其他用户，而不是follow这个用户的人，所以在建立邻接表的时候，要注意一下。对于每个输入的query，我们进行一次层序遍历（用队列实现），将每个用户对应的层数存在level数组中，大于题目所要求的的L后，break即可。
三、AC代码
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 1010;
vector<vector<int> > Adj;
int main()
{
int N, L, num, tmp, query;
scanf("%d%d", &N, &L);
Adj.resize(N+1);
for(int i=1; i<=N; i++)
{
scanf("%d", &num);
for(int j=0; j<num; j++)
{
scanf("%d", &tmp);
Adj[tmp].push_back(i);
}
}
scanf("%d", &num);
for(int i=0; i<num; i++)
{
int cnt = 0, level[N+1];
queue<int> q;
bool inq[maxn] = {
false};
scanf("%d", &query);
inq[query] = true;
q.push(query);
level[query] = 0;
while(!q.empty())
{
tmp = q.front();
q.pop();
if(level[tmp] >= L) break;
for(int i=0; i<Adj[tmp].size(); i++)
{
if(!inq[Adj[tmp][i]])
{
q.push(Adj[tmp][i]);
inq[Adj[tmp][i]] = true;
level[Adj[tmp][i]] = level[tmp] + 1;
cnt++;
}
}
}
printf("%d\n", cnt);
}
return 0;
}