HDU 1671 Phone List (字典树)

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15879    Accepted Submission(s): 5343

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
   
   
   
   

 

大体题意：

输入一些电话号码，问有没有一个号码是其他一个的前缀。

思路：

一开始想的想建立一个二维数组，记录所有电话号码，在建树，在查找，这样不仅占内存大，而且比较耗时。

后来经过学长的启发:

可以输入一个，直接查树（建树的时候v的值就改变一下，在建完一个单词后，在单词后面标记V=-1，这样可以根据v是否等于-1来进行判断！），不存在的话就插入到树里，否则继续，并且标记ok=false；

这样最后根据ok的值来进行输出。

一次循环，并且没有开二维数组，方法比较巧妙！

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
using namespace std;
const int maxn = 10;
char ss[50];
typedef struct Trie{
    Trie *next[maxn];
    int v;
}Trie;
Trie *root;
void creatTrie(char *str){
    Trie *p = root, *q;
    int len = strlen(str);
    for (int i = 0; i < len; ++i){
        int id = str[i] - '0';
        if (p->next[id] == NULL){
            q = new Trie;
            for (int i = 0; i < maxn; ++i)q->next[i] = NULL;
            q->v=1;
            p->next[id] = q;
            p=q;
        }else{
            p->next[id]->v++;
            p = p->next[id];
        }
    }
    p->v=-1;
}
int findTrie(char *str){
    int len = strlen(str);
    Trie *p = root;
    for (int i = 0; i < len; ++i){
        int id = str[i] - '0';
        if (p -> next[id] == NULL)return 0;
        if (p -> next[id]->v == -1)return 1;
        p=p->next[id];
    }
    return 1;
}
void deal(Trie *T){
    if (T == NULL)return;
    for (int i = 0; i < maxn; ++i)
        if (T -> next[i] != NULL)deal(T->next[i]);
    free(T);
}
int main()
{
    int n;
    cin >> n;
    while(n--){
        int m;
        cin >> m;
        root = new Trie;;
        for (int i = 0; i < maxn; ++i)root->next[i] = NULL;
        bool ok = true;
        for (int i = 0; i < m; ++i){
            scanf("%s",ss);
            if (findTrie(ss)){ok=false;continue;}
            creatTrie(ss);
        }
        if (ok)cout << "YES" << endl;
        else cout << "NO" <<endl;
        deal(root);
    }
    return 0;
}