POJ 3630 Phone List

Phone List

Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 3630
64-bit integer IO format: %lld      Java class name: Main
 

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

Source

Nordic 2007
 
解题：trie
 
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 struct trie {
18     int wd[10];
19     bool flag;
20     void init() {
21         memset(wd,-1,sizeof(wd));
22         flag = false;
23     }
24 };
25 trie dic[100000];
26 int tot,n;
27 char str[10010];
28 bool insertWd(int root,char *s) {
29     for(int i = 0; s[i]; i++) {
30         if(dic[root].flag) return false;
31         if(dic[root].wd[s[i]-'0'] == -1) {
32            dic[tot].init();
33             dic[root].wd[s[i]-'0'] = tot++;
34         }
35         root = dic[root].wd[s[i]-'0'];
36     }
37     for(int i = 0; i < 10; i++)
38         if(dic[root].wd[i] > 0) return false;
39    if(dic[root].flag) return false;
40     return dic[root].flag = true;
41 }
42 int main() {
43     int t,root = 0;
44     scanf("%d",&t);
45     while(t--) {
46         tot = 1;
47         dic[0].init();
48         scanf("%d",&n);
49         bool flag = true;
50         while(n--) {
51             scanf("%s",str);
52             if(flag) flag = insertWd(root,str);
53         }
54         flag?puts("YES"):puts("NO");
55     }
56     return 0;
57 }
View Code

 

posted @ 2014-08-30 18:42  狂徒归来  阅读(233)  评论(0)  编辑  收藏  举报