Phone List

Time Limit: 1000MS Memory Limit: 65536K

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

中文題意：
輸入一堆由數字組成字串，問是否有一串數字是另一串數字的字首，有的話輸出NO，沒有則輸出YES。

題解：
按資料來看，用字典樹來處理是一個很好的做法，在建字典樹的過程中，特殊判斷一下，最後輸出。

#include<iostream>
 

#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;
struct node{
    int ch[10];
    void init()
    {
        for (int i=0;i<10;i++)
        ch[i]=0;
    }
    bool pan()//判斷這是否是一個已有字串的終點
    {
        for (int i=0;i<10;i++)
        if (ch[i]!=0) return 1;
        return 0;
    }
}tr[1000005*12];
char s[1000005];
int p=0,tot=0,len,n,t,flag[1000005];
void add(int id)
{
    int now=0;
    for (int i=0;i<len;i++)
    {
        int tmp=s[i]-'0';

        if (now!=0 && flag[now]!=id && (tr[now].pan()==0))
        p=-1; 
        //判斷之前已經加入字典樹的那些數串是否是此次加入字串的字首
        //感覺自己想的這個方法有點sb可以去別人的部落格參考一下
        //有沒有更好的判斷方法
        if (!tr[now].ch[tmp])
        {
            tot++;
            tr[now].ch[tmp]=tot;
            tr[tot].init();
            flag[tot]=id;
        }
        now=tr[now].ch[tmp];
    }
    if (flag[now]!=id) p=-1;
}
int main()
{
    scanf("%d",&t);
    for (int w=1;w<=t;w++)
    {
        tot=0;
        p=0;
        tr[tot].init();
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
        {
            scanf("%s",s);
            len=strlen(s);
            if (p==0)
            add(i);
        }
        if (p==-1) printf("NO\n");
        else printf("YES\n");
    }
}

還有題上沒給字串的長度範圍，坑的我re兩次。。。。。