Phone List(字典树)

电话号码一致性检查

最新推荐文章于 2022-04-09 22:40:57 发布

原创最新推荐文章于 2022-04-09 22:40:57 发布 · 651 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#字典树

算法专栏收录该内容

17 篇文章

订阅专栏

本文介绍了一种使用Trie树算法来判断一组电话号码是否一致的方法。一致性指任一号码都不是另一号码的前缀，避免了拨号过程中的混淆。通过构建Trie树，算法在插入每个号码的同时检查是否出现前缀冲突。

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

Sample Output

NO
YES

题意

给定n个长度不超过10的数字串，问其中是否存在两个数字串S,T，使得S是T的前缀。有多组数据，数据组数不超过40.n<=10^4

思路

考虑将所有的字符串构成一棵Trie，在构建过程中可以顺便判断答案。若当前串插入后没有新建任何结点，则当前串肯定是之前插入的某个串的前缀；若插入过程中，有某个经过的结点带有串尾标记，则之前插入的某个串是当前串的前缀。依据上述两种情况判断

代码

#include<stdio.h>
#include<string.h>
#define N 100005
#define Z 10
int ch[N][Z],tot,bo[N];
int Insert(char *s) //插入一个字符串同时判断答案
{
    int len=strlen(s);
    int u=1,flag=0,i;
    for(i=0;i<len;i++)
    {
        int c=s[i]-'0';
        if(!ch[u][c])
            ch[u][c]=++tot;
        else if(i==len-1)
            flag=1;  //没有插入任何新结点
        u=ch[u][c];
        if(bo[u])
           flag=1;  //经过某个有标记的结点
    }
    bo[u]=1;
    return flag;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,ans=0;
        char s[20];
        scanf("%d",&n); tot=1;
        memset(ch,0,sizeof(ch));
        memset(bo,0,sizeof(bo));
        while(n--)
        {
            scanf("%s",s);
            if(Insert(s)) ans=1;
        }
        if(ans)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}