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[10];
    node()
    {
        count=0;
        int i;
        for(i=0; i<10; i++)
            child[i]=0;
    }
};

node *current, *newnode;

void insert(node *root, char *ss)
{
    //printf("%s*****\n",ss);
    int i, m;
    current=root;
    for(i=0; i<strlen(ss); i++)
    {
        m=ss[i]-'0';
        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=0; i<strlen(ss); i++)
    {
        m=ss[i]-'0';
        if(current->child[m]==NULL)
            return 0;
        current=current->child[m];
    }
    return current->count;
}

int main()
{
    char str[30], s[1010][30];
    int t, flag, i;
    while(scanf("%d",&t))
    {
        if(t==0) break;
        flag=0;
        node *root=new node;
        for(i=0; i<t; i++)
        {
            scanf("%s",str);
            strcpy(s[i], str);
            //puts(s[i]);
            insert(root, str);
        }
        for(i=0; i<t; i++)
        {
            if(search(root, s[i])-1)
            {
                flag=1;
                break;
            }
        }
        if(flag==0) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}