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最新推荐文章于 2025-07-26 13:10:24 发布

转载最新推荐文章于 2025-07-26 13:10:24 发布 · 63 阅读

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原文链接：http://www.cnblogs.com/ERKE/p/3259108.html

文章标签：

#java

本文介绍了一种用于检测电话号码列表一致性的算法，确保没有一个号码是另一个号码的前缀，通过字典树递归实现。适用于处理不超过10000个电话号码的场景。

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Phone List

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7343 Accepted Submission(s): 2525

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

字典树递归算法：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 #include <string.h>
 5 #include <math.h>
 6 #include <vector>
 7 #include <stack>
 8 #include <map>
 9 using namespace std;
10 #define ll long long int
11 #define INF 5100000
12 typedef struct node
13 {
14     struct node *next[10];
15     int n;
16     bool end;
17 } trie;
18 trie *inti()
19 {
20     trie *t;
21     t=(trie *)malloc(sizeof(trie));
22     t->end=0;
23     t->n=0;
24     for(int i=0; i<10; i++) t->next[i]=NULL;
25     return t;
26 }
27 int insert(trie *t,char a[])
28 {
29 
30     if(t->end)return 0;
31     if(*a=='\0')
32     {
33         if(t->n)return 0;
34         t->end=1;
35         return 1;
36     }
37     t->n++;
38     if(!t->next[*a-'0'])t->next[*a-'0']=inti();
39     return insert(t->next[*a-'0'],a+1);
40 }
41 void del(trie* t)
42 {
43     int i;
44     for(i=0; i<10; ++i)
45     {
46         if(t->next[i] != NULL)
47         {
48             del(t->next[i]);
49         }
50     }
51     free(t);
52 }
53 int main()
54 {
55     int n,m;
56     int i,j;
57     cin>>n;
58     char a[11];
59     for(i=0; i<n; i++)
60     {
61         trie *t;
62         t=inti();
63         cin>>m;
64         int fla=0;
65         for(j=0; j<m; j++)
66         {
67            scanf("%s",a);
68             if(!fla&&!insert(t,a))
69             fla=1;
70         }
71         if(!fla)
72         cout<<"YES"<<endl;
73         else cout<<"NO"<<endl;
74         del(t);
75     }
76 }