HDU1671 Phone List

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

2
3
911
97625999
91125426
5
113
12340
123440
12345
98346

 

Sample Output

NO
YES

 

Source

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

 

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

trie树做法：

 #include<bits/stdc++.h>

 using namespace std;

 int n, t;
 bool ans;
 const int MAXN = ;
 int g[MAXN][], f[MAXN], gx, w[MAXN];

 bool add (int u, string s, int x) {
     if (x >= s.length()) return ;
     w[u] = ;
     if (f[g[u][s[x] - '']]) {
         return ;
     }
     else {
         if (!g[u][s[x] - '']) g[u][s[x] - ''] = ++gx;
         if (x == s.length() - ) {
             if (w[g[u][s[x] - '']]) return ;
             f[g[u][s[x] - '']] = ;
             w[g[u][s[x] - '']] = ;
             return ;
         }
         else {
             return add(g[u][s[x] - ''], s, x + );
         }
     }
 }

 int main() {
     cin >> t;
     while (t--) {
         ans = ;
         cin >> n;
         memset(f, , sizeof(f));
         memset(g, , sizeof(g));
         memset(w, , sizeof(w));
         gx = ;
         while (n--) {
             string s;
             cin >> s;
             ans = min(ans, add(, s, ));
         }
         if (ans) {
             cout << "YES\n";
         }
         else {
             cout << "NO\n";
         }
     }
     return ;
 }

排序，每对相邻的暴力验证一下：

 #include<bits/stdc++.h>

 using namespace std;

 string s[];

 int main() {
     int t;
     cin >> t;
     while (t--) {
         int n;
         cin >> n;
         for (int i = ; i <= n; ++i) cin >> s[i];
         sort(s + , s +  + n);
         bool tf = ;
         for (int i = ; i <= n && tf; ++i) {
             if (s[i].length() > s[i - ].length()) {
                 bool x = ;
                 for (int j = ; j < s[i - ].length(); ++j) {
                     if (s[i - ][j] != s[i][j]) {
                         x = ;
                         break;
                     }
                 }
                 if (x) tf = ;
             }
         }
         if (tf) cout << "YES\n";
         else cout << "NO\n";
     }
     return ;
 }