【PAT甲级】1076 Forwards on Weibo（30 分）(图的遍历bfs)

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博客围绕微博转发量计算展开，已知每个用户关注的用户id列表和转发最多层数，需计算特定用户发微博的最大潜在转发量。采用BFS算法，用标记数组确保每个用户仅转发一次，通过层数限制扩散程度。

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题目链接
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:

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 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

题意：给出每个用户关注的用户id列表，和转发最多的层数，求一个id发了条微博最多会有多少个人转发

思路：bfs，每个用户[id]只能转发一个，可以设定一个标记数组，而层数限制了扩散程度也就是level

代码：

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define drep(i,n,a) for(int i=n;i>=a;i--)
#define mem(a,n) memset(a,n,sizeof(a))
#define lowbit(i) ((i)&(-i))
typedef long long ll;
typedef unsigned long long ull;
const ll INF=0x3f3f3f3f;
const double eps = 1e-6;
const int N = 1e3+5;

int n,L;
vector<vector<int> >g;
struct Node {
    int id,level;
};
bool vis[N];
int bfs(Node& now) {
    mem(vis,0);
    queue<Node>que;
    que.push(now);
    vis[now.id]=1;///标记
    int cnt=0;
    while(!que.empty()) {
        now=que.front();
        que.pop();
        int id=now.id;
        for(auto nextId: g[id]) {
            if(now.level>=L) continue;
            if(!vis[nextId]) {
                vis[nextId]=1;
                Node next= {nextId,now.level+1};
                que.push(next);
                cnt++;
            }
        }
    }
    return cnt;
}
int main() {
    scanf("%d%d",&n,&L);
    g.resize(n+1);
    int m;
    for(int i=1; i<=n; i++) {
        scanf("%d",&m);
        for(int j=0; j<m; j++) {
            int user;
            scanf("%d",&user);///关注的用户
            g[user].push_back(i);///user被id为i的用户关注，反向建立关系
        }
    }
    int k,id;
    scanf("%d",&k);
    while(k--) {
        scanf("%d",&id);
        Node tmp= {id,0};
        printf("%d\n",bfs(tmp));
    }
    return 0;
}