Phone List(HDOJ 1671)

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.

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

My Code

#include 
#include 

using namespace std;
struct Trie
{
    Trie *next[10];
    int v;
} * root;
int tot;
void creatTrie(string str) //建树
{
    Trie *p = root, *q;
    int l = str.length();
    for (int i = 0; i < l; i++)
    {
        int id = str[i] - '0';
        if (p->next[id] == NULL)
        {
            q = new Trie; //创建一个新节点
            memset(q->next, 0, sizeof(q->next));
            q->v = 1;
            p->next[id] = q;
        }
        p = p->next[id];
    }
    p->v = -1; //标记结束点
}
bool findTrie(string str) //找前缀
{
    Trie *p = root;
    int l = str.length();
    for (int i = 0; i < l; i++)
    {
        int id = str[i] - '0';
        if (i < l - 1 && p->next[id]->v == -1)
            return true;
        p = p->next[id];
    }
    return 0;
}
void del(Trie *p)
{
    for (int i = 0; i < 10; i++)
        if (p->next[i] != NULL)
            del(p->next[i]);
    delete p;
}
string s[10005];
int main()
{
    int t, n;
    cin >> t;
    while (t--)
    {
        root = new Trie; //创建根节点
        memset(root->next, 0, sizeof(root->next));
        root->v = 1;
        cin >> n;
        for (int i = 0; i < n; i++)
        {
            cin >> s[i];
            creatTrie(s[i]);
        }
        bool f = 0;
        for (int i = 0; i < n; i++)
            if (findTrie(s[i]))
            {
                cout << "NO" << endl;
                f = 1;
                break;
            }
        if (!f)
            cout << "YES" << endl;
        del(root);          //必须释放内存，否则MLE
    }
    return 0;
}