《ACM书中题目》X

• 原题

  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.

• 理解思路
  本题是查找一个数是否在另一个数的前边出现过，如果逐条查找，会造成超时。所以应该将数据先排序，然后在有序数列中依次查找一条数据的后一个。

• AC代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
    vector<string>v;
    string s;
    int n,i,j,k,m;
    cin>>n;
    for(i=0;i<n;i++)
    {
        v.clear();
        cin>>m;
        for(j=0;j<m;j++)
        {
            cin>>s;
            v.push_back(s);
        }
        sort(v.begin(),v.end());
        for(k=0;k<m-1;k++)
        {
            if(v[k+1].find(v[k])==0)
            {
                cout<<"NO"<<endl;
                goto RL;
            }
        }
        cout<<"YES"<<endl;
        RL:continue;
    }
}

• 总结
  要注意每次大循环对数组的初始化操作，否则只能成功第一组数据。
  从ACM书中学到”goto”操作，比开始用flag做标识更简洁易懂。