南阳理工ACM 162 Phone List

Phone List

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

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:

• Emergency 911
• Alice 97 625 999
• 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.

输入

The first line of input gives a single integer, 1 ≤ t ≤ 10, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 ≤ n ≤ 100000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.

输出

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

样例输入

2391197625999911254265113123401234401234598346

样例输出

NOYES

来源

POJ

package acm.ny;import java.util.Arrays;import java.util.Scanner;/** * 先按字符串从大到小排序，再建立字典树。 * @author Administrator * */public class Ny0163_20130608_2105 {private static Node root = null;private static String [] strs = null;public static void main(String[] args) {Scanner in = new Scanner(System.in);int cases, n;boolean isYes = true;cases = in.nextInt();while(cases-->0){root = new Node();n = in.nextInt();strs = new String[n];isYes = true;for(int i = 0; i < n; ++i){strs[i] = in.next();}Arrays.sort(strs);for(int i = 0; i < n; ++i){if(isYes){isYes = addToTree(strs[i]);}}if(isYes){System.out.println("YES");}else{System.out.println("NO");}}}private static boolean addToTree(String str){Node temp = root;for(int i = 0; i < str.length(); ++i){if(null == temp.child[str.charAt(i)-'0']){temp.child[str.charAt(i)-'0'] = new Node();}temp = temp.child[str.charAt(i)-'0'];if(temp.isLeaf){return false;}if(i == str.length() - 1){temp.isLeaf = true;}}return true;}}class Node{boolean isLeaf = false;Node [] child = new Node[10];}