sdutoj 2151 Phone Number

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：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 using namespace std;

 struct node
 {
     int count;
     node *child[];
     node()
     {
         count=;
         int i;
         for(i=; i<; i++)
             child[i]=;
     }
 };

 node *current, *newnode;

 void insert(node *root, char *ss)
 {
     //printf("%s*****\n",ss);
     int i, m;
     current=root;
     for(i=; i<strlen(ss); i++)
     {
         m=ss[i]-'';
         if(current->child[m]!=NULL)
         {
             current=current->child[m];
             ++(current->count);
         }
         else
         {
             newnode=new node;
             ++(newnode->count);
             current->child[m]=newnode;
             current=newnode;
         }
     }
 }

 int search(node *root, char *ss)
 {
     //printf("%s*****\n",ss);
     int i, m;
     current=root;
     for(i=; i<strlen(ss); i++)
     {
         m=ss[i]-'';
         if(current->child[m]==NULL)
             return ;
         current=current->child[m];
     }
     return current->count;
 }

 int main()
 {
     char str[], s[][];
     int t, flag, i;
     while(scanf("%d",&t))
     {
         if(t==) break;
         flag=;
         node *root=new node;
         for(i=; i<t; i++)
         {
             scanf("%s",str);
             strcpy(s[i], str);
             //puts(s[i]);
             insert(root, str);
         }
         for(i=; i<t; i++)
         {
             if(search(root, s[i])-)
             {
                 flag=;
                 break;
             }
         }
         if(flag==) printf("YES\n");
         else printf("NO\n");
     }
     return ;
 }