Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19302    Accepted Submission(s): 6529

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

建立字典树，每插入一个号码则将号码所在路径的count++，最后看如果有的号码末尾节点的count值不为1，则说明重复走过该路径，该号码为某号码的前缀。

注意一定要释放内存， 否则会爆内存。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3
 4 struct node{
 5     int count;
 6     node*next[10];
 7     node(){
 8         count=0;
 9         for(int i=0;i<10;i++){
10             next[i]=NULL;
11         }
12     }
13 };
14
15 node* root;
16 node* a[10010];
17 int k=0;
18
19 void insert(char str[15]){
20     int len=strlen(str);
21     node *p=root;
22     for(int i=0;i<len;i++){
23         if(p->next[str[i]-‘0‘]==NULL)
24         p->next[str[i]-‘0‘]=new node;
25         p=p->next[str[i]-‘0‘];
26         p->count++;
27     }
28     a[k++]=p;
29 }
30
31 int serch(int n){
32     for(int i=0;i<k;i++){
33         if(a[i]->count!=1)
34         return 1;
35     }
36     return 0;
37 }
38
39 void del(node *p){
40     if(p==0)
41     return ;
42     for(int i=0;i<10;i++){
43         del(p->next[i]);
44     }
45     delete p;//free(p);
46 }
47
48 int main(){
49     int t;
50     char str[15];
51     cin>>t;
52     while(t--){
53         root=new node;
54         int n;
55         k=0;
56         cin>>n;
57         for(int i=0;i<n;i++){
58             //gets(str);
59             scanf("%s",str);
60             insert(str);
61         }
62         if(serch(n))
63         cout<<"NO"<<endl;
64         else
65         cout<<"YES"<<endl;
66         del(root);
67     }
68     return 0;
69 }