【笨方法学PAT】1076 Forwards on Weibo （30 分）

一、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 triger, assuming that everyone who can view the initial post will forward it once, and that only Llevels 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

二、题目大意

给出每个用户关注的人的id，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。

三、考点

BFS

四、注意

1、使用DFS有2个没有通过，换成BFS，注意定义struct保存depth。

五、代码

#include
#include
#include
#define N 20010
using namespace std;

struct nodeb {
	int val;
	int layer;
};
int n, l;
int num;
bool visit[N] = { false };

vector> v;
int bfs(nodeb root) {
	fill(visit, visit + N, false);
	queue que;
	que.push(root);
	visit[root.val] = true;
	int cnt = 0;
	while (!que.empty()) {
		nodeb tmp = que.front();
		que.pop();
		int id = tmp.val;
		for (int i = 0; i < v[id].size(); ++i) {
			int index = v[id][i];
			if (visit[index] == false && tmp.layer < l) {
				que.push({ index,tmp.layer + 1 });
				visit[index] = true;
				cnt++;
			}
		}
	}
	return cnt;
}

int main() {
	//read
	cin >> n >> l;
	v.resize(n + 1);
	for (int i = 1; i <= n; ++i) {
		int k;
		cin >> k;
		for (int j = 0; j < k; ++j) {
			int a;
			cin >> a;
			v[a].push_back(i);
		}
	}

	//search
	int k;
	cin >> k;
	for (int i = 0; i < k; ++i) {
		fill(visit, visit + N, false);
		int index;
		cin >> index;
		/*num = 0;
		visit[index] = true;
		dfs(index, 1);
		cout << num << endl;*/
		int num = bfs(nodeb{ index,0 });
		cout << num << endl;
	}

	system("pause");
	return 0;
}