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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 23722 | Accepted: 7289 |
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:
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
题意是，给你N个数字，然后问你里面有没有一些数字是另外一些数字的前缀，如果是的话，就输出NO，不是的话就输出YES
唔，正版做法我觉得应该是字典树
但是这道题可以取巧，直接按照字典序大小，sort一下，然后比较a[i]和a[i+1]，看看a[i]是不是a[i+1]的前缀啥的~
string a[maxn]; int main() { int t; scanf("%d",&t); while(t--) { int n; cin>>n; vector<string> s; string ss; for(int i=0;i<n;i++) { cin>>ss; s.push_back(ss); } sort(s.begin(),s.end()); int flag=1; for(int i=0;i<n-1;i++) { if(s[i+1].find(s[i])==0) flag=0; } if(flag==0) cout<<"NO"<<endl; else cout<<"YES"<<endl; } }
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原文地址：http://www.cnblogs.com/qscqesze/p/4304015.html
