poj3630 Phone List

Phone List

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 31545		Accepted: 9244

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

Source

Nordic 2007

大致题意：给定n个字符串，问有没有两个字符串使得一个是另一个的前缀.

分析：Trie模板题，需要判断的情况有自己是不是其它串的前缀，有没有其它串是自己的前缀，对于第一个判断，如果在插入的过程中没有新建任何节点，就是其它串的前缀，对于第二个判断，如果中途遇到了结尾标记，就是的.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 1e5 + 5;

int T, n, len, tot, ans;
char s[1000010];

struct node
{
    int tr[30];
    bool end;
    void clear() {
        memset(tr, 0, sizeof(tr));
        end = false;
    }
}e[100010];

bool insert(char *ss)
{
    bool flagg = false;
    int u = 1, len = strlen(ss);
    for (int i = 0; i < len; i++)
    {
        if (!e[u].tr[ss[i] - '0'])
            e[u].tr[ss[i] - '0'] = ++tot;
        else
            if (i == len - 1)
                flagg = 1;
        if (e[u].end)
            flagg = 1;
        u = e[u].tr[ss[i] - '0'];
    }
    e[u].end = 1;
    return flagg;
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= tot; i++)
            e[i].clear();
        tot = 1;
        ans = 0;
        for (int i = 1; i <= n; i++)
        {
            scanf("%s", s);
            if (insert(s))
                ans = 1;
        }
        if (!ans)
            puts("YES");
        else
            puts("NO");
    }

    return 0;
}