1076 Forwards on Weibo

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 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

浅析

Q：在给定深度下，一个节点所能到达的其他节点数

一开始想用层数做为全局变量，但是 BFS 不知道何时到上一层的尾结点因此行不通。后改层数为每个节点各自的属性，源节点为 0，在遍历其邻接点时实现递增

Code

#include 
#include 
#include 
using namespace std;

const int maxn = 1010;
int L;

struct Node{
    int id;
    int level;
    Node(int _id,int _level):id(_id),level(_level){}
};
vector G[maxn];
int BFS(int s){
    bool inq[maxn] = {false};
    int num_share = 0;
    queue q;
    Node start = Node(s, 0);
    q.push(start);
    inq[start.id] = true;
    while(!q.empty()){
        Node temp = q.front();
        q.pop();
        for (int i = 0; i < G[temp.id].size();i++){
            Node next = G[temp.id][i];
            next.level = temp.level + 1;
            if(inq[next.id]==false&&next.level<=l){< span>                q.push(next);                inq[next.id] = true;                num_share++;            }        }    }    return num_share;}int main(){    int n;    scanf("%d%d", &n, &L);    for (int i = 0; i < n;i++){        int m;        scanf("%d", &m);        for (int j = 0; j < m;j++){            int u;            scanf("%d", &u);            G[u-1].push_back(Node(i, 0));   //巨坑！！题目的编号是从1开始的，因此要用u-1        }    }    int n_out;    scanf("%d", &n_out);    for (int i = 0; i < n_out;i++){        int s;        scanf("%d", &s);        printf("%d\n", BFS(s-1));      //s-1同上    }    return 0;}
                q.push(next);
                inq[next.id] = true;
                num_share++;
            }
        }
    }
    return num_share;
}
int main(){
    int n;
    scanf("%d%d", &n, &L);
    for (int i = 0; i < n;i++){
        int m;
        scanf("%d", &m);
        for (int j = 0; j < m;j++){
            int u;
            scanf("%d", &u);
            G[u-1].push_back(Node(i, 0));   //巨坑！！题目的编号是从1开始的，因此要用u-1
        }
    }
    int n_out;
    scanf("%d", &n_out);
    for (int i = 0; i < n_out;i++){
        int s;
        scanf("%d", &s);
        printf("%d\n", BFS(s-1));      //s-1同上
    }
    return 0;
}