首页 > 专题系列 > Java解POJ > POJ 3630 Phone List [解题报告] Java
2013
11-13

POJ 3630 Phone List [解题报告] Java

Phone List

问题描述 :

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 ≤ 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.

输出:

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

样例输入:

2
3
911
97625999
91125426
5
113
12340
123440
12345
98346

样例输出:

NO
YES

解题代码:

import java.util.Scanner;   
  
public class Main {   
    private static Trie trie = new Trie();   
    private static boolean isConsistent = true;   
  
    public static void main(String[] args) {   
        Scanner sc = new Scanner(System.in);   
        int t = sc.nextInt();   
        for (int i = 0; i < t; ++i) {   
            int n = sc.nextInt();   
            sc.nextLine();   
            trie = new Trie();   
            isConsistent = true;   
            for (int j = 0; j < n; ++j) {   
                String phone = sc.nextLine();   
                if (isConsistent)   
                    buildTrie(phone);   
            }   
            if (isConsistent)   
                System.out.println("YES");   
            else  
                System.out.println("NO");   
        }   
    }   
  
    static void buildTrie(String s) {   
        // char[] c = s.toCharArray();   
        int len = s.length();   
        Trie tmpTrie = trie;   
        for (int i = 0; i < len; ++i) {   
            int ch = Integer.parseInt(s.substring(i, i + 1));   
            // System.out.println(ch+"---"+(int)ch);   
            Trie tmp = tmpTrie.next[ch];   
            if (tmp == null) {   
                tmp = new Trie();   
                tmp.node = 1;   
                if (i == len - 1)   
                    tmp.isDispear = true;// 琛ㄧず杩��涓�釜�佃����灏�  
                tmpTrie.next[ch] = tmp;   
                tmpTrie = tmp;   
            } else {   
                if (tmp.isDispear) {   
                    isConsistent = false;   
                    break;   
                }   
  
                if (i == len - 1) {   
                    isConsistent = false;   
                    break;   
                }   
                tmpTrie = tmp;   
            }   
        }   
    }   
  
    private static class Trie {   
        int node = 0;   
        boolean isDispear = false;//������涓��璇��������涓�釜   
        Trie[] next = new Trie[10];   
    }   
}

  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. 为什么for循环找到的i一定是素数叻,而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak,而你每次取余都用的是原来的m,也就是n

  3. 你的理解应该是:即使主持人拿走一个箱子对结果没有影响。这样想,主持人拿走的箱子只是没有影响到你初始选择的那个箱子中有奖品的概率,但是改变了其余两个箱子的概率分布。由 1/3,1/3 变成了 0, 2/3