Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 29416 | Accepted: 8774 |
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
开内存池!
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<sstream> #include<algorithm> #include<queue> #include<deque> #include<vector> #include<cmath> #include<map> #include<stack> #include<set> #include<fstream> #include<memory> #include<list> #include<string> using namespace std; typedef long long LL; typedef unsigned long long ULL; #define MAXN 10003 #define LLL 1000000000 #define INF 1000000009 #define eps 0.00000001 /* 给定一些字符串,求这些字符串中是否存在一个字符串是另一个字符串的前缀 用trie tree解决 在插入时,分两种情况: 1.当前字符串的长度小于Tree中字符串长度 ,s[i] == ' '的情况 2.和当前字符串长度大于或等于Tree中字符串长度,在Tree中每个字符串结束的地方标记一下! */ bool flag,first;//这个标志是不是已经产生 上述情况 char s[MAXN]; typedef struct TreeNode { bool Endflag; struct TreeNode* Next[10]; }*Tree; TreeNode Memory[1000000]; int alloc = 0; Tree Newnode() { Tree T = &Memory[alloc++]; for (int i = 0; i < 26; i++) { //T->Next[i] = (Tree)malloc(sizeof(TreeNode)); T->Next[i] = NULL; T->Endflag = false; } return T; } Tree Insert(char s[], Tree t) { int p = 0; Tree T = t; while (s[p]) { int k = s[p] - '0'; if (!T->Next[k]) T->Next[k] = Newnode(); else { if (T->Next[k]->Endflag||(!T->Next[k]->Endflag&&s[p+1]=='