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最新推荐文章于 2020-09-20 09:46:38 发布

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本文介绍了一种使用排序和遍历的方法来判断一组电话号码是否互为前缀的问题，通过一个具体的算法实现，解决了电话号码簿中号码互斥性的判断。

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

题解：

自己做TLE两发。看别人都是用字典序，参考了其中一个vector+string中，发现其实STL sort排序其实就是字典序的，只需要排序，o(n)扫一下，在后一个字符串中查找前一个字符串是否出现过。

代码：

#include <iostream>
#include <map>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;


int main()
{
    vector<string> vec;
    int T;
    cin>>T;
    int n;
    string str;
    while(T--)
    {
      cin>>n;
      for(int i=0;i<n;i++)
      {
            cin>>str;
            vec.push_back(str);
      }
      sort(vec.begin(),vec.end());

      bool flag=true;
      string tmp,des;

      for(int i=0;i<n-1;i++){
          if(vec[i+1].find(vec[i])==0)
          {
              flag=false;
              break;
          }
      }
        vec.clear();
        if(flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;

    }
    return 0;
}