Phone List（字典树 or 排序）

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

解题思路：

用字典树做其实是做复杂了，每次还得释放内存空间，而且字典树代码也比较长，这题也可以用排序和比较函数来做。

字典树的代码在POJ上是TLE，在HDU上是AC。必须每次都得释放内存空间，否则：

POJ上交字典树：

HDU上交字典树：

排序 + 比较：

后者更高效。

AC代码：

字典树：

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int maxn = 10;
struct Trie
{
    int n;
    Trie *next[maxn];
};
Trie *root;

void init()
{
    root = (Trie *)malloc(sizeof(Trie));
    for(int i = 0; i < maxn; i++)
        root -> next[i] = NULL;
}

void insert(char *word)
{
    Trie *temp = root;
    while(*word != '\0')
    {
        if(temp -> next[*word - '0'] == NULL)
        {
            Trie *cur = (Trie *)malloc(sizeof(Trie));
            cur -> n = 1;
            for(int i = 0; i < maxn; i++)
                cur -> next[i] = NULL;
            temp -> next[*word - '0'] = cur;
        }
        else
            temp -> next[*word - '0'] -> n++;
        temp = temp -> next[*word-'0'];
        word++;
    }
    temp -> n = -1;
}

bool search(char *word)
{
    Trie *temp = root;
    for(int i = 0; word[i] != '\0'; i++)
    {
        temp = temp -> next[word[i] - '0'];
        if(temp == NULL)
           return false;
        if(temp -> n == -1)
            return true;
    }
    return true;
}

void del(Trie *cur)
{
    for(int i = 0; i < maxn; i++)
    {
        if(cur -> next[i] != NULL)
            del(cur -> next[i]);
    }
    free(cur);
}

int main()
{
    int t, n;
    char phone[15];
    scanf("%d", &t);
    while(t--)
    {
        init();
        bool flag = false;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%s", phone);
            if(search(phone))
                flag = true;
            if(flag)
                continue;
            insert(phone);
        }
        if(flag)
            printf("NO\n");
        else
            printf("YES\n");
        del(root);
    }
    return 0;
}

排序：

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

char phone[10000][11];
int cmp(const void* a,const void* b)
{
    return strcmp((char*)a,(char*)b);
}

int main(void)
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int i,n;
        bool flag=false;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%s",phone[i]);
        }
        qsort(phone,n,sizeof(phone[0]),cmp);
        for(i=0;i<n-1;i++)
        {
            if(strncmp(phone[i],phone[i+1],strlen(phone[i]))==0)
            {
                flag=true;
                break;
            }
        }
        if(flag==false)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}