PAT甲级1076 Forwards on Weibo (30 分)

题目：
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

思路：又错在读题上，完全读不懂题，以为题目要求存无向图双向边，实则是存单向边，因为题意是给出每个用户粉丝的数量，问如果该用户发了微博，会有多少用户转发。。

代码：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1005;

struct node {
	int v,d;
	node(int a,int b):v(a),d(b){}
};

vector <int> g[maxn];
int vis[maxn];
queue <node> q;
int dist[maxn];

int main() {
	int n,l;
	ios::sync_with_stdio(0);
	cin >> n >> l;
	for(int i = 1; i <= n; i++) {
		int t;
		cin >> t;
		while(t--) {
			int v;
			cin >> v;
			g[v].push_back(i); 
		}
	}
	int k;
	cin >> k;
	while(k--) {
		int u;
		cin >> u;
		for(int i = 1; i <= n; i++)vis[i] = dist[i] = 0;
		int cnt = 0;
		q.push(node(u,0));
		while(!q.empty()) {
			node p = q.front();q.pop();
			if(vis[p.v])continue;
			dist[p.v] = p.d;
			vis[p.v] = 1;
			if(p.d >= 1)cnt++;
			for(int i = 0; i < g[p.v].size(); i++) {
				int v = g[p.v][i];
				if(!vis[v] && p.d < l) {
					q.push(node(v,p.d+1));
				}
			}
		}
		cout << cnt << "\n";
	}
	return 0;
}