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Forwards on Weibo (30)

[编程题]Forwards on Weibo (30)

热度指数：5024 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

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.

输入描述:

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.

输出描述:

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.

示例1

输入

输出

4
5

算法知识视频讲解

Java

HackerLzh

测试点4超时了，不能用邻接矩阵存储吗？

发表于 2025-01-19 17:34:55 回复(0)

Java

米线宝宝他爹

没有使用深度优先遍历，用一个list保存每一个节点的追随者，先计算所有节点的第一层follows
然后在计算第二层follows，即在第一层的follow的基础上找下一层，依次类推。保存每一层新增的
follow

	import java.util.*;



public class Main {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();

        int L = scanner.nextInt();

        Map<Integer,Follows> followsMap = new HashMap<>();

        for (int i = 1; i <= n; i++){

            Set<Integer> set = new HashSet<>();

            followsMap.put(i,new Follows(set));

        }

          //初始化第一层

        for (int i = 1; i <= n; i++){

            int num = scanner.nextInt();

            for (int j = 0 ;j < num; j++){

                followsMap.get(scanner.nextInt()).follows.get(0).add(i);



            }

        }

     //计算每一层

        for(int l = 1; l < L; l++) {

            //计算每一个的下一层

for (int i = 1; i <= n; i++) {

                calculateNextLayer(followsMap, l, i);

            }

        }

        int m = scanner.nextInt();

     //输出

        for (int i = 0; i < m;i++){

            int index = scanner.nextInt();

            List<Set<Integer>> follows = followsMap.get(index).follows;

            Set<Integer> temp = new HashSet<>();

            for (int j =0; j< follows.size();j++){

                try {

                    temp.addAll(follows.get(j));

                }catch (Exception e){



                }

            }

            System.out.println(temp.size());

        }

    }



    private static void calculateNextLayer(Map<Integer,Follows> followsMap, int l, int num) {

        Follows follows = followsMap.get(num);

        //当前层

Set<Integer> current = new HashSet<>(follows.follows.get(0));

        Iterator<Integer> cur = new HashSet<>(current).iterator();

      //保存下一层   

Set<Integer> c = new HashSet<>();

        while (cur.hasNext()) {

            try {

                c.addAll(followsMap.get(cur.next()).follows.get(l - 1));

            }catch (Exception e){



            }

        }

        c.remove(num);

        follows.follows.add(c);

    }



    static class Follows{

        List<Set<Integer>> follows = new ArrayList<>();

        public Follows(Set<Integer> follows) {

            this.follows.add(follows);

        }

    }

}