HDU1671 字典树

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23480    Accepted Submission(s): 7893

Problem Description

Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let’s say the phone catalogue listed these numbers:
1. Emergency 911
2. Alice 97 625 999
3. Bob 91 12 54 26
In this case, it’s not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob’s phone number. So this list would not be consistent.

 

Input

The first line of input gives a single integer, 1 <= t <= 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 <= n <= 10000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.

 

Output

For each test case, output “YES” if the list is consistent, or “NO” otherwise.

 

Sample Input

2 3 911 97625999 91125426 5 113 12340 123440 12345 98346

 

Sample Output

NO YES 题意：给定一个数t t组数据 n代表下列有几个字符串 匹配是否一个字符串是否为另一个子串

题解: 构建字典树 利用字典树的性质去查看是否有子串 代码如下:

import java.util.Scanner;

class Node {
	int x;
	Node node[];

	public Node(int x, Node node[]) {
		// TODO Auto-generated constructor stub
		this.x = x;//记录是否为字符串的末尾
		this.node = node;
	}
}

public class Main {
	@SuppressWarnings("resource")
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		while (scanner.hasNext()) {
			int t = scanner.nextInt();
			for (int i = 0; i < t; i++) {
				int n = scanner.nextInt();
				Node root = new Node(0, new Node[10]);
				int x = 0;
				for (int j = 0; j < n; j++) {
					Node kNode = root;
					String string = scanner.next();
					int index = 0;
					while (index < string.length() && x == 0) {
						if (kNode.node[string.charAt(index) - '0'] == null) {
							if (kNode.x == 1) {
								x++;
							}
							if (index == string.length() - 1) {
								kNode.node[string.charAt(index) - '0'] = new Node(1, new Node[10]);
								kNode = kNode.node[string.charAt(index) - '0'];
								index++;
							} else {
								kNode.node[string.charAt(index) - '0'] = new Node(0, new Node[10]);
								kNode = kNode.node[string.charAt(index) - '0'];
								index++;
							}
						} else {
							if (index == string.length() - 1) {
								x++;
							}
							kNode = kNode.node[string.charAt(index) - '0'];
							index++;
						}
					}
				}
				if (x != 0) {
					System.out.println("NO");
				} else {
					System.out.println("YES");
				}
			}
		}
	}
}