HDU 1671 字典树

Phone List

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
  
  

这个题目与下面的poj1056是一模一样的..没有什么变化

#include <iostream>
using namespace std;
char s[10001][55];
int vis[10001];
struct node
{
	node *next[26];
	int flag;
	node()
	{
		memset(next,0,sizeof(next));
		flag=0;
	}
};
node *root=NULL;
void build(char *s)
{
	node *p=root;
	int i,len=strlen(s);
	for(i=0;i<len;i++)
	{
		if(p->next[s[i]-'0']==NULL)
			p->next[s[i]-'0']=new node;
		p=p->next[s[i]-'0'];
		p->flag++;
	}
}
int findstr(char *s)
{
	node *p=root;
	int i,len=strlen(s);
	for(i=0;i<len;i++)
	{
		if(p->next[s[i]-'0']!=NULL)
			p=p->next[s[i]-'0'];
		else
			return 0;
	}
	return p->flag;
}
void del(node *p)
{
	for(int i=0;i<10;i++)
		if(p->next[i])
			del(p->next[i]);
	delete p;
}
int main()
{

	int t,n,i;
	char s1[55];
	scanf("%d",&t);
	while(t--)
	{
		root=new node;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%s",s[i]);
			build(s[i]);
		}
		int judge=0;
		int len,j;
		for(i=0;i<n;i++)
		{
			if(findstr(s[i])>1)
			{
				judge=1;
				break;
			}
		}
		judge==0?cout<<"YES"<<endl:cout<<"NO"<<endl;
		del(root);
	}
	return 0;
}