poj3630 Phone List Dictionary Tree

Description

Given a list of phone numbers, determine if it is consistent in the sense this no number is the prefix of another. Let's say the phone catalogue listed these numbers:emergency 911 Alice 625 999 Bob 91 12 54 26

In this case, it's not possible-to-call Bob, because-the central would direct your-to-the-emergency line as soon as Y OU had dialled the first three digits of Bob ' s phone number. So the list would not being consistent.

Input

The first line of input gives a single integer, 1≤t≤40, the number of test cases. Each test case is starts with N, the number of the phone numbers, and on a separate line, 1≤n≤10000. Then follows n lines with a unique phone number on each line. A phone number is a sequence of at the most ten digits.

Output

For each test case, the 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

Try the dictionary tree, the first time with a dictionary tree, there are several places to pay attention to, are in the comments inside ...

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace
Std
	struct trie{int cnt;
struct trie* erza[10];
}node[100005];//open a little bit int num=0;
	Char number[100005][11];//open a bit trie* build () {trie* p=&node[num++];
	p->cnt=0;
for (int k=0;k<10;k++) p->erza[k]=null;//do not forget to initialize return p; 
} void Insert (Trie *&root,char *number)//must add & symbol .....
	{int i=0;
	if (root==null) root=build ();
	Trie* P=root;
		while (Number[i]) {if (!p->erza[number[i]-' 0 ']) p->erza[number[i]-' 0 ']=build ();
		p=p->erza[number[i]-' 0 '];
		p->cnt++;
	i++;
	}} bool PD (trie* root,char* number) {int i=0;
	if (root==null) return true;
	Trie* P=root;
		while (Number[i]) {p=p->erza[number[i]-' 0 '];
		if (p->cnt==1) return false;
	i++;
} return true;
} int n;
	int main () {int t,i,j;
	cin>>t;
	Trie *root;
		while (t--) {bool flag=true;
		cin>>n;
		Root=null;
			for (i=0;i<n;i++) {cin>>number[i]; InsertRoot,number[i]);
				} for (i=0;i<n;i++) {if (PD (Root,number[i))) {flag=false;
			Break
		}} if (flag) cout<< "yes\n";
		else cout<< "no\n";
	num=0; }
}