Phone List
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10837 Accepted Submission(s): 3735
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.
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
简单字典树：
代码： 实现 insert ,query,delete 三部分
1 /*hdu 1671 ×ÖµäÊ÷*/ 2 #define LOCAL 3 #include<cstdio> 4 #include<cstring> 5 #include<iostream> 6 using namespace std; 7 struct Trie 8 { 9 struct Trie *next[10]; 10 bool tail; 11 }; 12 char str[10005][12]; 13 void in_Trie(char *s,Trie *root) 14 { 15 Trie *cur=root,*newcur; 16 for(int i=0;s[i]!=' ';i++) 17 { 18 if(cur->next[s[i]-'0']==NULL) 19 { 20 newcur=new Trie; //(Trie*)malloc(sizeof(sizeof(Trie))); 21 for(int j=0;j<=9;j++) 22 newcur->next[j]=NULL; 23 newcur->tail=0; 24 cur->next[s[i]-'0']=newcur; 25 } 26 cur=cur->next[s[i]-'0']; 27 } 28 cur->tail=1; 29 } 30 bool query(char *s,Trie *root) 31 { 32 int i=0; 33 Trie *cur=root; 34 for(i=0;s[i];i++){ 35 if(cur->tail==0&&cur->next[s[i]-'0']!=NULL) 36 cur=cur->next[s[i]-'0']; 37 else break; 38 } 39 if(s[i]!=' ') return 1; 40 return 0; 41 } 42 void dele(Trie *root) 43 { 44 for(int i=0 ; i<=9 ; i++ ) 45 if(root->next[i]!=NULL) 46 dele(root->next[i]); 47 // free(root); 48 delete root; 49 } 50 int main() 51 { 52 #ifdef LOCAL 53 freopen("test.in","r",stdin); 54 #endif 55 int t,i,n; 56 Trie *root; 57 scanf("%d",&t); 58 while(t--) 59 { 60 root = new Trie ; 61 for(int j=0;j<=9;j++) 62 root->next[j]=NULL; 63 root->tail=0; 64 65 scanf("%d",&n); 66 for(i=0;i<n;i++){ 67 scanf("%s",str[i]); 68 in_Trie(str[i],root); 69 } 70 71 for(i=0;i<n;i++) 72 if(query(str[i],root)!=0) 73 break; 74 dele(root); 75 if(i>=n) 76 printf("YES "); 77 else 78 printf("NO "); 79 } 80 return 0; 81 }