HDU1671 Phone List

字典树水题，我在每一个节点都用了一个used来表示它是否被经过了，也用了一个end标志表示它是否为一个单词的结尾，虽然思路不是最好的，但是也AC了。。。。

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12405    Accepted Submission(s): 4209

Problem 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:
1. Emergency 911
2. Alice 97 625 999
3. 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
   
   
   
   

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
using namespace std;
typedef struct trie
{
    trie * next[10];
    bool isend;
    bool used;
};
int ans;
trie * newnode()
{
    trie * a = new trie;
    a->isend = false;
    a->used = false;

    for(int i = 0; i < 10; i++)
    {
        a->next[i] = NULL;
    }

    return a;
}
trie * Insert(trie * root, char a[])
{
    trie * use = root;

    for(int i = 0; a[i]; i++)
    {
        if(use->next[a[i] - '0'] == NULL)
        {
            use->next[a[i] - '0'] = newnode();
        }
        if(use->next[a[i] - '0']->isend==true)
            ans=0;
        if(i!=0)
            use->used=true;
        use = use->next[a[i] - '0'];
    }

    if(use->isend || use->used)
    {
        ans = 0;
    }

    use->isend = true;
    use->used = true;
    return root;
}
void Clear(trie * root)
{
    for(int i = 0; i < 10; i++)
    {
        if(root->next[i] != NULL)
        {
            Clear(root->next[i]);
        }
    }

    delete root;
}
int main()
{
//    freopen("D://input.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        trie* root=newnode();
        ans=1;
        int n;
        scanf("%d",&n);
        char number[12];
        for(int i=0;i<n;i++)
        {
            scanf("%s",number);
            Insert(root,number);
        }
        if(ans)
            printf("YES\n");
        else printf("NO\n");
        Clear(root);
    }
    return 0;
}