hdu 1671 Phone List（字典树）

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.

 

题意：判断输入的一系列号码中有没有出现一个号码是另一个号码的前缀，如果有输出NO否则YES

 

明显是一道字典树的问题，可以直接套模版

 

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int M = 1e4 + 10;
char gg[20];
struct TnT {
    TnT *next[15];
    int vis;
    TnT():vis(0){memset(next,0,sizeof(next));}
}*cmp[M];
int k;
void build(TnT *root , char s[]) {
    int len = strlen(s);
    TnT *p = root;
    for(int i = 0 ; i < len ; i++) {
        int id = s[i] - '0';
        if(p->next[id] == NULL) {
            p->next[id] = new TnT;
        }
        p = p->next[id];
        p->vis++;
    }
    cmp[k++] = p;
}
int find() {
    for(int i = 0 ; i < k ; i++) {
        if(cmp[i]->vis != 1) {
            return 1;
        }
    }
    return 0;
}
void de(TnT *root) {
    if(root == NULL)
        return ;
    for(int i = 0 ; i < 12 ; i++) {
        de(root->next[i]);
    }
    delete root;
}
int main()
{
    int t;
    cin >> t;
    TnT *root;
    while(t--) {
        int n;
        cin >> n;
        root = new TnT;
        k = 0;
        for(int i = 0 ; i < n ; i++) {
            cin >> gg;
            build(root , gg);
        }
        if(find()) {
            cout << "NO" << endl;
        }
        else {
            cout << "YES" << endl;
        }
        de(root);
    }
    return 0;
}