POJ 3630-Phone List【比较许多字符中有没有一个是另一个的前缀，简单方法】

青旅半醒 2022-07-24 07:19 48阅读 0赞

Phone List

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;26766</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;8055</td> 
  </tr> 
 </tbody> 
</table>

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:

*  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][]

#include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<string>
    #define N 10010
    using namespace std;
    string tel[N];
     
    bool judge(int n)
    {
        int i,k,len;
        for(k=0;k<n-1;k++)
        {
            len=tel[k].size();//<tel[k+1].size()?tel[k].size():tel[k+1].size();
            for(i=0;i<len;i++) 
                if(tel[k][i]!=tel[k+1][i])
                    break;
            if(i==len) return true;
        }
        return false;
    }
     
    int main()
    {
        int t,i,n;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            for(i=0;i<n;i++) cin>>tel[i];
            sort(tel,tel+n);
            if(judge(n)) printf("NO\n");
            else printf("YES\n");
        }
        return 0;
    }

[Nordic 2007]: http://poj.org/searchproblem?field=source&key=Nordic+2007