HDU 1671 Phone List（字典树）

最新推荐文章于 2020-05-09 10:07:54 发布

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本文介绍了一个电话号码列表的一致性检查问题，通过构建字典树判断列表中的电话号码是否互为前缀，提供了完整的代码实现。

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Phone List

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25761 Accepted Submission(s): 8610

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.

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

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

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题意：给定n个电话号码串，问这n个电话号码串中是否存在某一串是其它串的前缀，如果存在输出NO，否则YES。

思路：在建立字典树的时候判断一下当前串是不是其他串的前缀，或者其他串是不是它的前缀即可。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int trie[maxn][10],tot,vis[maxn];
int inst(char s[])
{
    int len=strlen(s);
    int root=0;
    int flag=0;
    for(int i=0;i<len;i++)
    {
        int id=s[i]-'0';
        if(!trie[root][id])
        {
            trie[root][id]=++tot;
            flag=1;
        }
        root=trie[root][id];
        if(vis[root]) return 1;
    }
    vis[root]=1;
    if(!flag) return 1;
    return 0;
}

int main()
{
    int t,n;
    char s[10005];
    scanf("%d",&t);
    int flag;
    while(t--)
    {
        flag=0,tot=0;
        memset(trie,0,sizeof(trie));
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%s",s);
            int ans=inst(s);
            if(ans) flag=1;
        }
        if(flag) printf("NO\n");
        else printf("YES\n");
    }
}