1004 Counting Leaves

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0, the number of nodes in a tree, and M (<), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

import java.util.* ;

public class Main{
    static class Node{
        int id;
        Queue<Node> children;
        Node(int id){
            this.id=id;
        }
    }
    static HashMap<Integer,Node> tree;
    static int N;//树的节点个数
    static int M;//有孩子的节点数
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        M = sc.nextInt();
        tree = new HashMap<>();
        for(int i=;i<M;i++){
            int id=sc.nextInt();
            int K = sc.nextInt();
            Node node = new Node(id);
            Queue<Node> children = new LinkedList<>();
            for(int j=;j<K;j++){
                Node no = new Node(sc.nextInt());
                children.offer(no);
            }
            node.children=children;
            tree.put(id,node);
        }
        Queue<Node> que =new LinkedList<>();
        que.add(tree.get());
        BFS(que);
    }

    public static void BFS(Queue<Node> que){
        int len = ;
        Queue<Node> NextQue = new LinkedList<>();
        while(que.size()!=){
            Node node = que.poll();
            if(tree.size()==||!tree.containsKey(node.id)){
                len++;
            }else{
                NextQue.addAll(tree.get(node.id).children);
            }
        }
        if(NextQue.size()==){
            System.out.print(len);
        }
        else {
            System.out.print(len+" ");
            BFS(NextQue);
        }
    }
}

思路为BFS。

HashMap建树。