Phone list

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

#include
#include
#include
#include
#include
#include
#include
using namespace std;
struct node
{
    node *next[10]; 
    int flag;
    node()
    {
        int i;
        for(i=0;i<10;i++)
        {
            next[i]=NULL;
        }
        flag=0;   ///如果为0，表示这个节点并不是尾结点。
    }
};
node *root;
node *current;
void del(node *root)           ///我所不懂的是为什么不写这个会出现memeory limited exceed?
{
    for(int i=0;i<10;i++)
    {
        if(root->next[i]!=NULL)
        {
            del(root->next[i]);
        }
    }
    free(root);
    return;
}
void insert1(char *st)
{
    int id,i;
    current=root;
    for(i=0;inext[id]==NULL)
        {
            current->next[id]=new node;
            current=current->next[id];
        }
        else
        {
            current=current->next[id];
        }
    }
    current->flag=1;
}
int find1(char *sr)
{
    current=root;int len=strlen(sr);
    int id,i;
    for(i=0;inext[id]!=NULL)        
        {
            current=current->next[id];
            if(current->flag==1) ///表示这个节点是某个电话号码的末尾号，说明到这个节点的前缀是一个电话。
            {
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int t,n,i;
    scanf("%d",&t);
    while(t--)
    {
        root=new node;
        char tel[10000][11];
        scanf("%d",&n);
        for(i=0;i

还有更简单的方法，在插入时就可以进行判断了。