# [poj 3630] Phone List trie 树 Phone List
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 26448 Accepted: 7972

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

SOL： 动态建树； 建时1.路径上有flag=1的点；
2.建后还有路径；

``````#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int n,flag;
int tot;
char s;
struct node{
int ch;
int flag;
void init()
{
flag=0;
memset(ch,0,sizeof(ch));
}
}t[10005*11];

void insert(char s[])
{
int now=0;
for(int i=0;s[i];i++)
{
int tmp=s[i]-'0';
if(!t[now].ch[tmp])
{
t[now].ch[tmp]=++tot;
t[tot].init();
}
now=t[now].ch[tmp];
if(t[now].flag)  flag=1;
}
for(int i=1;i<=9;i++)
if(t[now].ch[i])
flag=1;
t[now].flag=1;
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
tot=0;
t.init();
flag=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%s",s);
insert(s);
}
if(flag==1) printf("NO\n");
else printf("YES\n");
}
}``````
原文作者：Trie树
原文地址: https://blog.csdn.net/ALPS233/article/details/50765637
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