Phone List (prefix judgment-two-point lookup)

Description Given A list of phone numbers, determine if it is consistent in the sense this no number is the prefix of Anot Her. 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-to-the-emergency line as soon as Y OU had dialled the first three digits of Bob ' s phone number. So the list would not being consistent.

Input the first line of input gives a single integer, 1 <= t <=, the number of test cases. Each test case is starts with N, the number of the phone numbers, on a separate line, 1 <= n <= 10000. Then follows n lines with a unique phone number on each line. A phone number is a sequence of at the 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

Main topic:

give a few sets of strings to determine if there is a string in each set of strings is the prefix of another string, if so, output no, if not, output yes.

Input:

The first line, an integer t (1<=t<=40), indicates how many sets of test data, the next T-group data, the first row of each group of data is an integer n (1<=n<=10000), which represents the number of data in the group, followed by a string of n rows per line.

Problem Solving Ideas:

The subject can be solved with a dictionary tree, but I use a simple lookup. For each set of data, the string is sorted by dictionary order, and then only by comparing each of the two adjacent strings to determine if they have a prefix relationship.

The code is as follows: C + + code

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
using namespace std;
String s[10005];
Determines whether the adjacent two strings after sorting is a string that is the prefix of another string
bool judge (int n)
{
    int i,j,m;
    for (i=0;i<n-1;i++)
    {
        m=s[i].size ();
        for (j=0;j<m;j++)
        {
            if (S[i][j]!=s[i+1][j]) break
                ;
        }
        if (j==m)
            return false;
    }
    return true;
}
int main ()
{
    int t,n,i;
    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d", &n);
        for (i=0;i<n;i++)
            cin>>s[i];//can only be entered with CIN, not scanf input
        sort (s,s+n);//Sort by Dictionary
        if (judge (n))
           printf ("yes\n");
        else
           printf ("no\n");
    }
    return 0;
}