E

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：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=91073#problem/E

题目意思就是说输入的号码中不能有任意一个号码是另一个号码的前缀

思路：所有号码按照字典序进行排序，（sort函数可以实现字典排序），然后遍历所有数，看前一个数是否是后一个数的前缀

注意：

定义string s[100];

输入若写 scanf("%s", s[i]);就错了，提示

strlen(s[i])也用不了，会提示

因为string类型不是单纯的字符串，不能用%s表示，而strlen里面应该放的是字符串的名字，

输入改成 cin>>s[i];就对了

求长度要用s.length()，或者s.size(), 一定一定要记得加括号，要不然又有下面的提示，std怎么着……

对于string类型也不能用sizeof(s[0]);这样的话无论怎样，输出结果都是一样的。

如果对定义的字符串数组string s[100]按照字典序排序，可以用sort(s, s+n);

c++里面定义string变量用的应该是#include<string>而且要用using namespace std；

代码

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