HDOJ 1671 Phone List

Description

Given 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:

• Emergency 911
• Alice 97 625 999
• 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.

Input

The 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.

Output

For each test case, output "YES" if the list is consistent, or "NO" otherwise.

Sample Input

2
3
911
97625999
91125426
5
113
12340
123440
12345
98346

Sample Output

NO
YES

  这道题就是判断一组字符串中是否有一个字符串是另一个字符串的前缀.

 

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
int k;
typedef structnode
{
    int num;
    structnode *next[10];
}node;
nodememory[1000000];

int insert(char *s, node*T)
{
    int len = 0;
    int id = 0;
    node*p,*q;
    p = T;
    len = strlen(s);
    for(int i = 0 ; i < len; i++)
    {
        id = s[i] - '0';
        if(p->num == 1)
          return 1;
        if(p->next[id] == NULL)
        {
            q = &memory[k++];
            q->num = 0;
            for(int j = 0; j < 10; j++)
            {
                q->next[j] = NULL;
            }

            p->next[id] = q;
        }
        p = p->next[id];
    }

    for(int i = 0 ; i < 10; i++)
    {
        if(p->next[i] != NULL)
           return 1;
    }
    p->num = 1;
    return 0;
}
int main()
{
    int t;
    node*T;

    cin>>t;
    while(t)
    {
        k = 0;
        T = &memory[k++];
        T->num = 0;
        for(int i = 0 ; i < 10; i++)
        {
            T->next[i] = NULL;
        }
        int flag = 0;
        int n;
        cin>>n;
        for(int i = 0 ; i < n ; i++)
        {
            char s[15];
            cin>>s;
            if(flag) 
              continue;
            if(insert(s,T))
              flag = 1;
        }
        if(flag)
          cout<<"NO"<<endl;
        else
          cout<<"YES"<<endl;
        t--;
    }
    return 0;
}