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
复制
样例输出：
NO YES
这道题可以选用字典树的方法，感觉就像是二叉树的一种衍生吧，多记一下。
#include<iostream>
#include<string.h>
using namespace std;
struct tree
{
int val;
struct tree* T[10];
tree(){
val=0;
memset(T,NULL,sizeof(T));
}
};
typedef struct tree tRee;
typedef tRee* Tree;
Tree root;
bool create(char a[])
{
Tree p=root;
int length=strlen(a);
bool flag=true;
for(int n=0;n<length;n++)
{
int k=a[n]-'0';
if(p->T[k]==NULL)
{
flag=false;
p->T[k]=new tRee;
}
p=p->T[k];
if(p->val)
{
return false;
}
}
p->val=1;
if(flag)return false;
else return true;
}
int main()
{
int sum;
cin>>sum;
for(int n=0;n<sum;n++)
{
int number;
bool flag=true;
cin>>number;
root=new tRee;
for(int m=0;m<number;m++)
{
char a[10];
cin>>a;
if(flag){
flag=create(a);
}
}
if(flag)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
