Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10321    Accepted Submission(s): 3543


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這題用了兩種方法，一種是把他們排序，再相鄰兩個進行比較，如果發現有包涵關係立馬退出，例外一種是用字典樹進行匹配

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef struct Trie{
   int v;
   Trie *next[10];
}Trie;
Trie *root;
bool flag;
void inti()//初始化
{
    root = (Trie *)malloc(sizeof(Trie));
    root->v=0;
    for(int i=0;i<10;i++)
        root->next[i]=NULL;
}
void Create(char *str)//插入
{
    int len=strlen(str);
    Trie *p = root,*q;
    for(int i=0;i<len;i++)
    {
        int id=str[i]-'0';
        if(p->next[id]==NULL)
        {
            q = (Trie *)malloc(sizeof(Trie));
            q -> v = 0;
            for(int i=0;i<10;i++)
            {
                q->next[i]=NULL;
            }
            p->next[id]=q;

        }
        else
        {
            if(p->next[id]->v==1||str[i+1]=='\0')//表示字典樹中包涵這個串的子串
            {
                flag=true;
                return ;
            }
        }
       p=p->next[id];
    }
    p->v=1;
}
void Shifang(Trie *p)//釋放記憶體
{
    if(p==NULL)return ;
    for(int i=0;i<10;i++)
    {
        if(p->next[i]!=NULL)
            Shifang(p->next[i]);
    }
    free(p);
    return ;
}
int main()
{
   char str[100];
   int t,n;

   scanf("%d",&t);
   while(t--)
   {    inti();
       flag=false;
       scanf("%d",&n);
       for(int i=0;i<n;i++)
       {
           scanf("%s",&str);
           if(!flag)Create(str);
       }
       if(!flag)printf("YES\n");
       else printf("NO\n");
       Shifang(root);//必須加這個，不然會爆的
   }
   return 0;
}
 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
string s[10004];
int n;
int fun()
{
    int i,j;
    for(i=1;i<n;i++)
    {
        int len=s[i-1].length();
        for(j=0;j<len;j++)
        {
            if(s[i][j]!=s[i-1][j])break;
        }
        if(j>=len)return false;
    }
    return true;
}
int main()
{
   int t;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%d",&n);
       for(int i=0;i<n;i++)
         cin>>s[i];
       sort(s,s+n);
       if(fun())
        cout<<"YES"<<endl;
       else cout<<"NO"<<endl;
   }
   return 0;
}