http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2151
Phone Number
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描述
We know that if a phone number A is another phone number B’s prefix, B is not able to be called. For an example, A is 123 while B is 12345, after pressing 123, we call A, and not able to call B.
Given N phone numbers, your task is to find whether there exits two numbers A and B that A is B’s prefix.
输入
The input consists of several test cases.
The first line of input in each test case contains one integer N (0<N<1001), represent the number of phone numbers.
The next line contains N integers, describing the phone numbers.
The last case is followed by a line containing one zero.
输出
For each test case, if there exits a phone number that cannot be called, print “NO”, otherwise print “YES” instead.
示例输入
2 012 012345 2 12 012345 0
示例输出
NO YES
提示
来源
2010年山东省第一届ACM大学生程序设计竞赛
示例程序
AC代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 using namespace std;
6
7 struct node
8 {
9 int count;
10 node *child[10];
11 node()
12 {
13 count=0;
14 int i;
15 for(i=0; i<10; i++)
16 child[i]=0;
17 }
18 };
19
20 node *current, *newnode;
21
22 void insert(node *root, char *ss)
23 {
24 //printf("%s*****\n",ss);
25 int i, m;
26 current=root;
27 for(i=0; i<strlen(ss); i++)
28 {
29 m=ss[i]-‘0‘;
30 if(current->child[m]!=NULL)
31 {
32 current=current->child[m];
33 ++(current->count);
34 }
35 else
36 {
37 newnode=new node;
38 ++(newnode->count);
39 current->child[m]=newnode;
40 current=newnode;
41 }
42 }
43 }
44
45 int search(node *root, char *ss)
46 {
47 //printf("%s*****\n",ss);
48 int i, m;
49 current=root;
50 for(i=0; i<strlen(ss); i++)
51 {
52 m=ss[i]-‘0‘;
53 if(current->child[m]==NULL)
54 return 0;
55 current=current->child[m];
56 }
57 return current->count;
58 }
59
60 int main()
61 {
62 char str[30], s[1010][30];
63 int t, flag, i;
64 while(scanf("%d",&t))
65 {
66 if(t==0) break;
67 flag=0;
68 node *root=new node;
69 for(i=0; i<t; i++)
70 {
71 scanf("%s",str);
72 strcpy(s[i], str);
73 //puts(s[i]);
74 insert(root, str);
75 }
76 for(i=0; i<t; i++)
77 {
78 if(search(root, s[i])-1)
79 {
80 flag=1;
81 break;
82 }
83 }
84 if(flag==0) printf("YES\n");
85 else printf("NO\n");
86 }
87 return 0;
88 }