Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10720    Accepted Submission(s): 3701

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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字符串排序再扫描一遍即可

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<stdlib.h>
 6 #include<algorithm>
 7 using namespace std;
 8 const int MAXN=10000+5;
 9 int T,n;
10 struct node
11 {
12     char str[10+5];
13 }a[MAXN];
14
15 int cmp(const node A,const node B)
16 {
17     return strcmp(A.str,B.str)<0;
18 }
19
20 int judge(char a[],char b[])
21 {
22     int len=strlen(a);
23     for(int i=0;i<len;i++)
24         if(a[i]!=b[i])
25             return 0;
26     return 1;
27 }
28 int main()
29 {
30     //freopen("in.txt","r",stdin);
31     scanf("%d",&T);
32     while(T--)
33     {
34         scanf("%d",&n);
35         for(int i=0;i<n;i++)
36             scanf("%s",a[i].str);
37
38         sort(a,a+n,cmp);//利用结构体对字符串排序
39
40         int ok=0;
41         for(int i=0;i<n-1;i++)
42         {
43             if(judge(a[i].str,a[i+1].str))
44             {
45                 ok=1;
46                 break;
47             }
48         }
49
50         if(ok)
51             printf("NO\n");
52         else
53             printf("YES\n");
54     }
55     return 0;
56 }