Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 18217    Accepted Submission(s): 6120

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

 

题目链接：

嗯还是一道简单的Trie，但是对于内存的要求比较高= =不管是malloc或new出来的，用完一定要释放不然……MLE数次。说是Trie其实是学习一下如何进行有效率地释放……

注意题目中可能给的单词顺序不同会出现两种存在前缀的情况

 

例1、输入911 后输入911000，这个比较简单，逐一插入911000的时候若节点存在则检查是否此时的节点的flag为true即这个节点是一个结尾点。

例2、输入911000 后输入911，办法有很多，我是用一个any来记录后面的911插入时是否出现过未出现的字母，显然在节点不存在new的时候any就要变成1了。

 

代码：

#include 
     
      #include 
      
       using namespace std;const int N=10;const int M=15;struct Trie{    Trie *nxt[N];    bool flag;    Trie()    {        for (int i=0; i
       
        nxt[indx])        {            Trie *one=new Trie();            any=true;            cur->nxt[indx]=one;            cur=one;        }        else        {            cur=cur->nxt[indx];            if(cur->flag)//过程中遇到结尾单词，前缀存在                is_end=true;        }    }    if(!any)        is_end=true;    cur->flag=true;    return is_end;}void deleTrie(Trie *L){    Trie *cur=L;    for (int i=0; i
        
         nxt[i])            deleTrie(cur->nxt[i]);    delete cur;}char s[M];int main(void){    int tcase,n;    scanf("%d",&tcase);    while (tcase--)    {        bool flag=true;        L=new Trie();        scanf("%d",&n);        while (n--)        {            scanf("%s",s);            if(flag)            {                if(update(s))                    flag=false;            }        }        puts(flag?"YES":"NO");        deleTrie(L);    }    return 0;}