Hdoj 1671

原题链接

描述

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.

输入

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.

输出

For each test case, output “YES” if the list is consistent, or “NO” otherwise.

样例输入

2

3

911

97625999

91125426

5

113

12340

123440

12345

98346

样例输出

NO

YES

思路

字典树

令人悲伤，数组形式的字典树MLE了，把数组大小改小又WA，原以为只能用链表了，群里大佬说能用vector形式！

我以前没用过vector，赶紧百度学了一下，上手套模版，稍微改动一下就好。期间犯了很多很傻的错误，比如忘记给vector迭代下去之类的，觉得自己好渣啊~

代码

#include <bits/stdc++.h>
#define maxn 900000
using namespace std;

struct node
{
    int a[10];
};

vector<node> trie;
int cnt[maxn], tot, f;

void create(char st[])
{
    int len = strlen(st);
    int u = 0;
    for(int i = 0; i < len; i++)
    {
        if(trie[u].a[st[i] - ‘0‘] == 0)
        {
            trie[u].a[st[i] - ‘0‘] = ++tot;
            node pnew;
            for(int j = 0; j <10; j++) pnew.a[j] = 0;
            trie.push_back(pnew);
        }
        u = trie[u].a[st[i] - ‘0‘];
        if(cnt[u]) f = 1;
    }
    cnt[u] = 1;
    for(int i = 0; i < 10; i++)
        if(trie[u].a[i]) f = 1;
}

int main()
{
    int num; scanf("%d", &num);
    while(num--)
    {
        trie.clear();
        memset(cnt, 0, sizeof(cnt));
        f = 0; tot = 0;
        node pnew; for(int j = 0; j <10; j++) pnew.a[j] = 0;
        trie.push_back(pnew);
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            char st[12]; scanf("%s", st);
            if(f == 0) create(st);
        }
        if(f == 0) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

原文地址：https://www.cnblogs.com/HackHarry/p/8379377.html