【PAT】A1076 Forwards on Weibo (30分)（图的遍历bfs）

1076 Forwards on Weibo (30分)

题目链接

法一：邻接矩阵法

#include<iostream>
#include<queue>
using namespace std;

const int maxn = 1010;

int Graph[maxn][maxn] = {0};
int n, l; 

//邻接矩阵版本 
struct Node{
	int id;
	int depth;
	Node(int _id, int _depth):id(_id),depth(_depth){}
};

int bfs(int u)
{
	int numPerson = 0;
	queue<Node> q;
	bool vis[maxn] = {false};
	struct Node node(u, 0);
	q.push(node);
	vis[u] = true;
	while(!q.empty())
	{
		struct Node temp = q.front();
		q.pop();
		if(temp.depth >= l)
			continue;
		for(int i=1;i<=n;i++)
		{
			if(!vis[i] && Graph[temp.id][i] != 0)
			{
				vis[i] = true;
				numPerson++;
				struct Node node(i, temp.depth+1);
				q.push(node);
			}
		}
	}
	return numPerson;
}

int main()
{
	std::ios::sync_with_stdio(false);
	cin>>n>>l;
	int counts, temp;
	for(int i=1;i<=n;i++)
	{
		cin>>counts;
		for(int j=0;j<counts;j++)
		{
			cin>>temp;
			Graph[temp][i] = 1;
		}
	}
	cin>>counts;
	int query;
	for(int i=0;i<counts;i++)
	{
		cin>>query;
		cout<<bfs(query)<<endl;
	}
	return 0;
}

法二：邻接表法

#include<iostream>
#include<queue>
using namespace std;

const int maxn = 1010;

//邻接表版本 
struct Node{
	int id;
	int depth;
	Node(int _id, int _depth):id(_id),depth(_depth){}
};

vector<int> adj[maxn];//邻接表 
int n, l; 

int bfs(int u)
{
	int numPerson = 0;
	queue<Node> q;
	bool vis[maxn] = {false};
	struct Node node(u, 0);
	q.push(node);
	vis[u] = true;
	while(!q.empty())
	{
		struct Node temp = q.front();
		q.pop();
		if(temp.depth >= l)
			continue;
		for(int i=0;i<adj[temp.id].size();i++)
		{
			if(!vis[adj[temp.id][i]])
			{
				vis[adj[temp.id][i]] = true;
				numPerson++;
				struct Node node(adj[temp.id][i], temp.depth+1);
				q.push(node);
			}
		}
	}
	return numPerson;
}

int main()
{
	std::ios::sync_with_stdio(false);
	cin>>n>>l;
	int counts, temp;
	for(int i=1;i<=n;i++)
	{
		cin>>counts;
		for(int j=0;j<counts;j++)
		{
			cin>>temp;
			adj[temp].push_back(i);
		}
	}
	cin>>counts;
	int query;
	for(int i=0;i<counts;i++)
	{
		cin>>query;
		cout<<bfs(query)<<endl;
	}
	return 0;
}