PAT 甲级 1076 Forwards on Weibo

最新推荐文章于 2021-08-18 19:29:53 发布

原创最新推荐文章于 2021-08-18 19:29:53 发布 · 436 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT甲级 #Forwards on Weibo #1076

PAT甲级/乙级机试经验同时被 2 个专栏收录

254 篇文章

订阅专栏

PAT甲级

159 篇文章

订阅专栏

本文介绍了一个基于图遍历算法的问题，旨在计算微博用户在特定层级内的最大转发潜力。通过使用BFS而非DFS避免了超时问题，确保了算法效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1076 Forwards on Weibo （30 point(s)）

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 L levels of indirect followers are counted.

Sample Input:

Sample Output:

4
5

经验总结：

这一题，原理上不是很难，就是图的遍历，不过使用方法上，建议使用BFS，用DFS最后一个测试点会超时，还有就是，注意一下，超过规定的层次无需再向下遍历~

AC代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
const int maxn=1010;
vector<int> adj[maxn];
int n,L,amount[maxn],num;
bool flag[maxn];
void DFS(int s,int level)
{
	if(level>=L)
		return ;
	for(int i=0;i<adj[s].size();++i)
	{
		if(flag[adj[s][i]]==false)
		{
			++num;
			flag[adj[s][i]]=true;
		}
		DFS(adj[s][i],level+1);
	}
}

int main()
{
	int m,k,t;
	memset(amount,-1,sizeof(amount));
	scanf("%d%d",&n,&L);
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&m);
		for(int j=0;j<m;++j)
		{
			scanf("%d",&t);
			adj[t].push_back(i);
		}
	}
	scanf("%d",&k);
	for(int i=0;i<k;++i)
	{
		scanf("%d",&t);
		if(amount[t]!=-1)
			printf("%d\n",amount[t]);
		else
		{
			memset(flag,0,sizeof(flag));
			num=0;
			flag[t]=true;
			DFS(t,0);
			amount[t]=num;
			printf("%d\n",num);
		}
	}
	return 0;
}