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本文作者：ljh2000
作者博客：http://www.cnblogs.com/ljh2000-jump/
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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:
- 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.
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 正解：Trie解题报告： 1A就是爽。 询问有没有一个串是另一个串的前缀。 建一棵Trie树，然后每次check一下，当前单词的最后一个结点是不是之前已经存在，存在则说明是另一个串的前缀； 如果中间经过了某个结点是之前的某个串的结束结点，也说明存在前缀。 打个标记即可。
//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int MAXN = 100011;
int n,tr[MAXN][10],cnt;
bool ok,vis[MAXN];
char ch[12];
inline int getint(){
int w=0,q=0; char c=getchar(); while((c<‘0‘||c>‘9‘) && c!=‘-‘) c=getchar();
if(c==‘-‘) q=1,c=getchar(); while (c>=‘0‘&&c<=‘9‘) w=w*10+c-‘0‘,c=getchar(); return q?-w:w;
}
inline void read(){
if(!ok) { scanf("%s",ch); return ; }
char c=getchar(); int u=0,now; bool flag;
while(c>‘9‘ || c<‘0‘) c=getchar();
while(1) {
now=c-‘0‘; if(vis[u]) ok=false;
if(!tr[u][now]) tr[u][now]=++cnt,flag=false;
else flag=true;
u=tr[u][now];
c=getchar(); if(c<‘0‘||c>‘9‘) break;
}
if(vis[u] || flag) { ok=false; return ; }
vis[u]=1;
}
inline void work(){
int T=getint();
while(T--) {
memset(tr,0,sizeof(tr)); memset(vis,0,sizeof(vis));
cnt=0; ok=true;
n=getint(); for(int i=1;i<=n;i++) read();
if(ok) puts("YES"); else puts("NO");
}
}
int main()
{
work();
return 0;
}