Phone List

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem 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:
1. Emergency 911
2. Alice 97 625 999
3. 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

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

Recommend

lcy | We have carefully selected several similar problems for you: 1075 1247 1800 1677 1298

Statistic | Submit | Discuss | Note

http://acm.hdu.edu.cn/showproblem.php?pid=1671

trie树做法：

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int n, t;
 6 bool ans;
 7 const int MAXN = 110000;
 8 int g[MAXN][15], f[MAXN], gx, w[MAXN];
 9 
10 bool add (int u, string s, int x) {
11     if (x >= s.length()) return 0;
12     w[u] = 1;
13     if (f[g[u][s[x] - '0']]) {
14         return 0;
15     }
16     else {
17         if (!g[u][s[x] - '0']) g[u][s[x] - '0'] = ++gx;
18         if (x == s.length() - 1) {
19             if (w[g[u][s[x] - '0']]) return 0;
20             f[g[u][s[x] - '0']] = 1;
21             w[g[u][s[x] - '0']] = 1;
22             return 1;
23         }
24         else {
25             return add(g[u][s[x] - '0'], s, x + 1);
26         }
27     }
28 }
29 
30 int main() {
31     cin >> t;
32     while (t--) {
33         ans = 1;
34         cin >> n;
35         memset(f, 0, sizeof(f));
36         memset(g, 0, sizeof(g));
37         memset(w, 0, sizeof(w));        
38         gx = 1;
39         while (n--) {
40             string s;
41             cin >> s;
42             ans = min(ans, add(1, s, 0));
43         }
44         if (ans) {
45             cout << "YES\n";
46         }
47         else {
48             cout << "NO\n";
49         }
50     }
51     return 0;
52 }

View Code

排序，每对相邻的暴力验证一下：

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 string s[10001];
 6 
 7 int main() {
 8     int t;
 9     cin >> t;
10     while (t--) {
11         int n;
12         cin >> n;
13         for (int i = 1; i <= n; ++i) cin >> s[i];
14         sort(s + 1, s + 1 + n);
15         bool tf = 1;
16         for (int i = 2; i <= n && tf; ++i) {
17             if (s[i].length() > s[i - 1].length()) {
18                 bool x = 1;
19                 for (int j = 0; j < s[i - 1].length(); ++j) {
20                     if (s[i - 1][j] != s[i][j]) {
21                         x = 0;
22                         break;
23                     }
24                 }
25                 if (x) tf = 0;
26             }
27         }
28         if (tf) cout << "YES\n";
29         else cout << "NO\n";
30     }
31     return 0; 
32 }

View Code

posted on 2018-05-21 10:44 ^m 阅读(204) 评论(0) 编辑收藏举报