首页 > 代码库 > HDU-1671
HDU-1671
Phone List
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19302 Accepted Submission(s): 6529Problem DescriptionGiven 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.
InputThe 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.
OutputFor each test case, output “YES” if the list is consistent, or “NO” otherwise.
Sample Input2391197625999911254265113123401234401234598346
Sample OutputNOYES
建立字典树,每插入一个号码则将号码所在路径的count++,最后看如果有的号码末尾节点的count值不为1,则说明重复走过该路径,该号码为某号码的前缀。
注意一定要释放内存, 否则会爆内存。
AC代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 struct node{ 5 int count; 6 node*next[10]; 7 node(){ 8 count=0; 9 for(int i=0;i<10;i++){ 10 next[i]=NULL; 11 } 12 } 13 }; 14 15 node* root; 16 node* a[10010]; 17 int k=0; 18 19 void insert(char str[15]){ 20 int len=strlen(str); 21 node *p=root; 22 for(int i=0;i<len;i++){ 23 if(p->next[str[i]-‘0‘]==NULL) 24 p->next[str[i]-‘0‘]=new node; 25 p=p->next[str[i]-‘0‘]; 26 p->count++; 27 } 28 a[k++]=p; 29 } 30 31 int serch(int n){ 32 for(int i=0;i<k;i++){ 33 if(a[i]->count!=1) 34 return 1; 35 } 36 return 0; 37 } 38 39 void del(node *p){ 40 if(p==0) 41 return ; 42 for(int i=0;i<10;i++){ 43 del(p->next[i]); 44 } 45 delete p;//free(p); 46 } 47 48 int main(){ 49 int t; 50 char str[15]; 51 cin>>t; 52 while(t--){ 53 root=new node; 54 int n; 55 k=0; 56 cin>>n; 57 for(int i=0;i<n;i++){ 58 //gets(str); 59 scanf("%s",str); 60 insert(str); 61 } 62 if(serch(n)) 63 cout<<"NO"<<endl; 64 else 65 cout<<"YES"<<endl; 66 del(root); 67 } 68 return 0; 69 }
HDU-1671
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。