(pat)A1076 Forwards on Weibo

本文介绍了一个模拟微博用户间转发行为的算法。通过构建社交网络图,计算特定用户在限定层级内的最大转发潜力。采用广度优先搜索策略遍历图结构,统计有效转发数。

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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 are 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:
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
这题我有个 问题,为什么在在队列里面放结构体的指针会,oj给错呢?
ac代码:

#include<iostream>
#include<cstdio>
#include<map> 
#include<vector>
#include<algorithm> 
#include<queue>
#include<cstring>
using namespace std;

vector<int> graph[1005];
int n,l;
struct node
{
    int id;
    int level;
};
char vis[1005];

int solve(int x)
{
    memset(vis,0,sizeof(vis));
    vis[x]=1;
    queue<node> que;
    node begin;
    begin.id=x;
    begin.level=0;
    que.push(begin);
    int countt=0;
    while(!que.empty())
    {
        node temp=que.front();
        que.pop();
        int now=temp.id;
        int lev=temp.level;
        //printf("%d\n",now);
        for(int i=0;i<graph[now].size();i++)
        {
            int u=graph[now][i];
            if(vis[u])
                continue;
            vis[u]=1;
            node h;
            h.id=u;
            h.level=lev+1;
            if(lev<l)
            {
                countt++;
                que.push(h);
            }
        }

    }
    return countt;
}
int main()
{
    scanf("%d%d",&n,&l);
    int num;
    int x;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&num);
        while(num--)
        {
            scanf("%d",&x);
            graph[x].push_back(i);  
        }
    }
    scanf("%d",&num);
    while(num--)
    {
        scanf("%d",&x);
        printf("%d\n",solve(x));
    }
    return 0;
}

wa:

#include<iostream>
#include<cstdio>
#include<map> 
#include<vector>
#include<algorithm> 
#include<queue>
#include<cstring>
using namespace std;

vector<int> graph[1005];
int n,l;
struct node
{
  int id;
  int level;
};
char vis[1005];

int solve(int x)
{
  memset(vis,0,sizeof(vis));
  vis[x]=1;
  queue<node*> que;//只有这里有变化
  node begin;
  begin.id=x;
  begin.level=0;
  que.push(&begin);
  int countt=0;
  while(!que.empty())
  {
    node* temp=que.front();
    que.pop();
    int now=temp->id;
    int lev=temp->level;
    //printf("%d\n",now);
    for(int i=0;i<graph[now].size();i++)
    {
      int u=graph[now][i];
      if(vis[u])
        continue;
      vis[u]=1;
      node h;
      h.id=u;
      h.level=lev+1;
      if(lev<l)
      {
        countt++;
        que.push(&h);
      }
    }

  }
  return countt;
}
int main()
{
  scanf("%d%d",&n,&l);
  int num;
  int x;
  for(int i=1;i<=n;i++)
  {
    scanf("%d",&num);
    while(num--)
    {
      scanf("%d",&x);
      graph[x].push_back(i);  
    }
  }
  scanf("%d",&num);
  while(num--)
  {
    scanf("%d",&x);
    printf("%d\n",solve(x));
  }
  return 0;
}
### Max-Forwards HTTP Header Purpose and Usage The `Max-Forwards` header field in HTTP requests is used to limit the number of times a request can be forwarded through proxies or gateways before reaching its destination server. Each time a proxy forwards the request, it decrements this value by one. When the value reaches zero, the recipient should not forward the message but handle it directly or respond with an appropriate status code indicating that forwarding has been exhausted. This mechanism helps prevent infinite loops when multiple intermediary servers are involved in processing a single request. It also provides insight into how many intermediaries have processed the request so far[^1]. #### Example Usage Scenario Consider a scenario where client A sends a TRACE method request intended ultimately for origin server Z via several intermediate proxies B, C, D: ```http TRACE /path/to/resource HTTP/1.1 Host: www.example.com Max-Forwards: 2 ``` In this case: - Proxy B receives the request first and sees `Max-Forwards: 2`. After handling, B reduces the count to 1. - Next, proxy C gets the modified request (`Max-Forwards: 1`) from B and again decreases the counter to 0 after processing. - Finally, since D now sees `Max-Forwards: 0`, D does not pass along the request further down the chain but responds back up toward the original sender instead. Such behavior ensures efficient routing while avoiding potential issues arising out of endless redirection chains among network nodes. --related questions-- 1. What happens if Max-Forwards exceeds initial setting during transmission? 2. Can all types of HTTP methods utilize the Max-Forwards directive effectively? 3. How do different web browsers interpret varying values provided within Max-Forwards headers? 4. Are there any security implications associated with manipulating Max-Forwards settings?
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