Phone Number(字典树)

【Description
Description】
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.
Input
【Input
Input】
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.
Output
【Output
Output】
For each test case, if there exits a phone number that cannot be called, print “NO”,
otherwise print “YES” instead.
Sample Input】
【Sample Input
2
012
012345
2
12
012345
0
Sample Output】
【Sample Output
NO
YES

#include<stdio.h>
#include<stdlib.h>
struct node
{
    int flag;
    struct node *next[10];
}*root;
int tag;
struct node *creat()
{
    struct node *p;
    p=(struct node *)malloc(sizeof(struct node));
    p->flag=0;
    for(int i=0;i<10;i++)
        p->next[i]=NULL;
     return p;
}
void inser(char *s)
{
    int i;
    struct node *p;
    p=root;
    for(i=0;s[i]!='\0';i++)
    {
        if(!(p->next[s[i]-'0'])) p->next[s[i]-'0']=creat();
        p=p->next[s[i]-'0'];
        if(p->flag==1) tag=1;判断p节点是否是某字符串的结束。
    }
    p->flag=1;
}
int main ()
{
    int t;
    char s[1010];
    while(~scanf("%d",&t))
    {
        if(!t) break;
        root=creat();
        tag=0;
        while(t--)
        {
            scanf("%s",s);
            inser(s);
        }
        if(tag==1)
                printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}