POJ3630 Phone List

Phone List

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28093		Accepted: 8408

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

Sample Output

NO
YES

Source

Nordic 2007

把所有的串插入trie树，对每个串的结尾进行标记。

然后进行检查，如果一个串包含了其他串的结尾结点，说明会冲突，输出NO

翻记录看到四个月前做过这道题，WA地飞起然后放弃了。

想着这次要飞速刷完，结果因为数据范围看漏了一个0，途中RE了两次……

 1 /**/
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 const int mxn=200050;
 9 struct node{
10     int c[15];
11     int tag;
12 }t[mxn];
13 char s[mxn][15];
14 int cnt;
15 int n;
16 bool flag;
17 void insert(char s[]){
18     int len=strlen(s);
19     int now=0,i,j;
20     for(i=0;i<len;i++){
21         int tmp=s[i]-'0';
22         if(!t[now].c[tmp])
23             t[now].c[tmp]=++cnt;
24         now=t[now].c[tmp];
25     }
26     t[now].tag++;
27     return;
28 }
29 void pd(char s[]){
30     if(flag)return;
31     int i,j;
32     int now=0,len=strlen(s);
33     for(i=0;i<len;i++){
34         int tmp=s[i]-'0';
35         
36         if(t[now].tag){
37             flag=1;
38             return;
39         }
40         now=t[now].c[tmp];
41     }
42     return;
43 }
44 int main(){
45     int T;
46     scanf("%d",&T);
47     int i,j;
48     while(T--){
49         memset(t,0,sizeof t);
50         cnt=0;
51         scanf("%d",&n);
52         for(i=1;i<=n;i++){
53             scanf("%s",s[i]);
54             insert(s[i]);
55         }
56         flag=0;
57         for(i=1;i<=n;i++)
58             pd(s[i]);
59         if(flag)printf("NO
");
60         else printf("YES
");
61     }
62     return 0;
63 }