A1076 Forwards on Weibo (BFS)

这道题是计算weibo最大转发次数，使用BFS方法。

#include <iostream>
#include <string>
#include <set>
#include <map>
#include<queue>
#include<stack>
#include<algorithm>

//new and delete
//A1076 Forwards on Weibo
//这里结点从0编号到N-1
using namespace std;

const int MaxN = 2010;
struct Node {
	int v;	//编号
	int layer=0;		//层号
};
//被关注图，邻接表
vector<Node> G[MaxN];
bool visit[MaxN] = { false };

//N (≤1000), the number of users; 
//and L (≤6), the number of levels of indirect followers that are counted
int N, L;	
void PrintG() {
	cout << "结点";
	for (int i = 0; i < N; i++)
	{
		cout << endl;
		for (int j = 0; j < G[i].size(); j++)
			cout << G[i][j].v<<" ";
	}
	cout << endl;
}
//将所有结点层数变成0
void iniLayer() {
	for (int i = 0; i < N; i++)
	{
		for (int j = 0; j < G[i].size(); j++)
			G[i][j].layer = 0;
	}
}
//BFS，返回转发数
int BFS(int x) {
	queue< Node > q;
	Node n;
	n.v = x; n.layer = 0;
	q.push(n);
	visit[x] = true;

	int num = -1;	//转发数，-1是去掉自己转发
	while (!q.empty()) {
		//访问
		n = q.front();
		q.pop();
		num++;

		//后继结点入队
		for (int i = 0; i < G[n.v].size(); i++) {
			G[n.v][i].layer = n.layer + 1;
			int N_v = G[n.v][i].v, N_l = G[n.v][i].layer;
			if (visit[N_v] == false && N_l <= L) {
				visit[N_v] = true;
				q.push(G[n.v][i]);
			}
		}
	}

	return num;
}
int main() {
	//Input data
	cin >> N >> L;
	for (int i = 0; i < N; i++) {
		//输入是i关注的一群人
		int j;
		cin >> j;
		for (int k = 0; k < j; k++) {
			Node temp;
			int a;
			cin >> a;
			a--;
			temp.v = i;
			G[a].push_back(temp);
		}

	}
	vector<int> query_ind;
	int i;
	cin >> i;
	for (int j = 0; j < i; j++) {
		int temp;
		cin >> temp;
		query_ind.push_back(--temp);
	}
	//
	//PrintG();
	//

	//output
	for (int i = 0; i < query_ind.size(); i++) {
		cout << BFS(query_ind[i]) << endl;
		iniLayer();
		fill(visit, visit + N,false);
	}


	return 0;

}