Phone List(poj3630,字典树)

/*http://poj.org/problem?id=3630
Phone List
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 18141 Accepted: 5773
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:


Emergency 911
Alice 97 625 999
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
Source


Nordic 2007
题意：问是否存在一个字符串是另一个字符串的前缀。
要点： 利用字典树的性质边构造字典树，边查找；
  注意事项：1，当建立新的节点时一定要清空空间
            2，一下出现这两种情况都应该退出
              a:自己已经是别的串的前缀了；
              b:别的串已经成了自己的前缀了
     
*/

#include<stdio.h>
#include<string.h>
#include <iostream>
using namespace std;
const int maxn=100000+10;
int trie[maxn][10];// trie[i][j]表示用来保存节点i的那个编号为j的字节点
int node[maxn];//表示当前节点的权值，当node[i]=1时形成一个字符，以节点i结尾
int cnt;
void init()
{
           cnt=1;
           memset(trie[0],0,sizeof(trie[0]));
}
int insert(char s[])//边插入，变查找
{
           int u=0;
           int len=strlen(s);
           for(int i=0;i<len;i++)
          {int v=s[i]-'0';
          if(trie[u][v]==0)  //1开辟新空间
            {memset(trie[cnt],0,sizeof(trie[cnt]));
            node[cnt]=0;
            trie[u][v]=cnt++;
            }
            u=trie[u][v];
            if(node[u]==1)//2已出现前缀
             return 1;
             if(i==len-1)
             {for(int j=0;j<10;j++)
                      if(trie[u][v]) //3成为别串的前缀
                      return 1;
             }


          }
          node[u]=1;//该串结束的标志
          return 0;


}




int main()
{
  int T,n,i;
   bool ok;
   char ch[12];
   scanf("%d",&T);
   while(T--)
  {
             init();
           ok=false;
            scanf("%d",&n);
           for( i=0;i<n;i++)
           {scanf("%s",ch);
           ok=insert(ch);
           if(ok==true) //一旦发现出现前缀退出
            break;
           }
           while(i<n-1)
           {scanf("%s",ch);
           i++;
           }


           if(ok==false)
           printf("YES\n");
           else
           printf("NO\n");
   }
   return 0;
}