POJ 3630 Trie树 TLE

动态 Trie 会超时，DISCUSS 中有人提醒要用静态数组的 Trie，还没学；

这个题要注意除了判断当前串是否有和字典中的串相同的前缀外，还要判断当前串是否是字典中串的前缀；

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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

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# include <stdio.h>

struct tree{
    tree *node[10];
    int bj;
    tree() {
        bj = 1;
        for (int i = 0; i < 10; ++i)
            node[i] = NULL;
    }
} *root;

int n;
char tel[15], ok;

void visit(tree *p, char *s)
{
    char c = (*s) - '0';
    if (p->node[c])
    {
        p = p->node[c];
        if (!s[1] || p->bj == 0) {ok = 0; return ;}
    }
    else
    {
        p->node[c] = new tree;
        p = p->node[c];
        p->bj = 2;
    }
    if (s[1]) visit(p, s+1);
    else p->bj = 0;
}

void release(tree *p)
{
    for (int i = 0; i < 10; ++i)
        if (p->node[i]) release(p->node[i]);
       delete p;
}

void solve()
{    
    ok = 1;
    root = new tree;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%s", tel);
        if(ok) visit(root, tel);      
    }
    puts(ok ? "YES":"NO");
    release(root);
}

int main()
{
    int T;
    
    scanf("%d", &T);
    while (T--)
    {
        solve();
    }
    
    return 0;
}

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