PAT 1076 Forwards on Weibo [图的遍历] [BFS] [邻接表/邻接矩阵]

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 Nlines 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

---------------------------------------这是题目和解题的分割线---------------------------------------

 输入是反着来的，题目提供的是关注的情况而不是被关注的情况。累加给定层数以内的粉丝，BFS的思想比较搭。这道题用邻接矩阵也不会超时，这里给出邻接矩阵和邻接表两种写法。

//邻接表 
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>

using namespace std;

int N,L,vis[1010],cnt;
vector<int> v[1010]; //用可变数组作为邻接表 

//由于这道题涉及层数，而BFS不进行递归，所以另开一个结构体记录 
struct node
{
	int id;
	int layer;
};
 
void bfs(int index)
{
	node tmp;
	tmp.id = index;
	tmp.layer = 0;
	queue<node> q;
	q.push(tmp);
	vis[index] = 1; //入队就标记 
	while(!q.empty())
	{
		node topNode = q.front();
		q.pop();
		int top = topNode.id;
		//在top的粉丝数范围内遍历 
		for(int i=0;i<v[top].size();i++)
		{
			node next;
			next.id = v[top][i];
			next.layer = topNode.layer+1;//层数在上一层基础上+1 
			//如果该粉丝没有遍历过，且层数满足要求 
			if(!vis[v[top][i]]&&next.layer<=L)
			{			
				q.push(next);
				vis[next.id] = 1;
				cnt++; //符合数+1 
			}
		}
	}
}

int main()
{
	int i,j,x,num;
	scanf("%d%d",&N,&L);
	//注意从1开始而不是0 
	for(i=1;i<=N;i++)
	{
		scanf("%d",&x);
		for(j=0;j<x;j++)
		{
			scanf("%d",&num);
			v[num].push_back(i); //输入时反过来，即i是num的粉丝 
		}
	}
	int k,qy;
	scanf("%d",&k);
	for(i=0;i<k;i++)
	{
		//每次统计人数都需要清除上一次标记的痕迹 
		memset(vis,0,sizeof(vis));
		cnt = 0; //每次遍历前置零  
		scanf("%d",&qy);
		bfs(qy);
		printf("%d\n",cnt);
	}
	return 0;
}

//邻接矩阵 
#include<cstdio>
#include<cstring>
#include<queue>

#define maxN 1010

using namespace std;

int N,L,G[maxN][maxN] = {},vis[maxN];

struct node
{
	int layer;
	int id;
	//构造函数，省一些些代码 
	node(int _id,int _layer)
	{
		id = _id;
		layer = _layer;
	}
};

int bst(int index)
{
	int count = 0;
	queue<node> q; 
	q.push(node(index,0));
	vis[index] = 1;
	while(!q.empty())
	{
		node topNode = q.front();
		q.pop();
		int top = topNode.id;
		for(int i=1;i<=N;i++)
		{
			node next = node(i,topNode.layer+1);
			//邻接矩阵要多加一个边是否存在的判断 
			if(G[top][i]&&!vis[i]&&next.layer<=L)
			{
				q.push(next);
				vis[i] = 1;
				count++;
			}
		}
	}
	return count;
}

int main()
{
	int i,j,x,num;
	scanf("%d%d",&N,&L);
	for(i=1;i<=N;i++)
	{
		scanf("%d",&x);
		for(j=0;j<x;j++)
		{
			scanf("%d",&num);
			G[num][i] = 1;
		}
	}
	int k,qy;
	scanf("%d",&k);
	for(i=0;i<k;i++)
	{
		scanf("%d",&qy);
		memset(vis,0,sizeof(vis)); //头文件时string.h 
		int cnt = bst(qy);
		printf("%d\n",cnt);
	}
	return 0;
}