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=0;i<M;i++){
- int id=sc.nextInt();
- int K = sc.nextInt();
- Node node = new Node(id);
- Queue<Node> children = new LinkedList<>();
- for(int j=0;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(1));
- BFS(que);
- }
- public static void BFS(Queue<Node> que){
- int len = 0;
- Queue<Node> NextQue = new LinkedList<>();
- while(que.size()!=0){
- Node node = que.poll();
- if(tree.size()==0||!tree.containsKey(node.id)){
- len++;
- }else{
- NextQue.addAll(tree.get(node.id).children);
- }
- }
- if(NextQue.size()==0){
- System.out.print(len);
- }
- else {
- System.out.print(len+" ");
- BFS(NextQue);
- }
- }
- }
思路为 BFS.
HashMap 建树.
来源: http://www.bubuko.com/infodetail-3067725.html