POJ 3630

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

题意：

给你 $n$ 个字符串，让你确定其中是否有存在某个字符串是另一个字符串的前缀。

AC代码：

（通过OpenJ_Bailian - 3791和OpenJ_Bailian - 4089）

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e4+5;

int n;
string s[maxn];

namespace Trie
{
    const int SIZE=maxn*10;
    int sz;
    struct TrieNode{
        int ed;
        int nxt[10];
    }trie[SIZE];
    void init()
    {
        sz=1;
        memset(trie,0,sizeof(trie));
    }
    void insert(const string& s)
    {
        int p=1;
        for(int i=0;i<s.size();i++)
        {
            int ch=s[i]-'0';
            if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
            p=trie[p].nxt[ch];
        }
        trie[p].ed++;
    }
    int prefix(const string& s)
    {
        int res=0,p=1;
        for(int i=0;i<s.size();i++)
        {
            p=trie[p].nxt[s[i]-'0'];
            res+=trie[p].ed;
            if(!p) return res;
        }
        return res;
    }
};

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        Trie::init();
        for(int i=1;i<=n;i++) {
            cin>>s[i];
            Trie::insert(s[i]);
        }

        bool ok=1;
        for(int i=1;i<=n;i++) {
            if(Trie::prefix(s[i])>1) {
                ok=0;
                break;
            }
        }
        if(ok) cout<<"YES
";
        else cout<<"NO
";
    }
}