杭电OJ 1671解题报告（字典树模板）

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14925    Accepted Submission(s): 5029

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

/*
题目大意是你打电话 如果某个号码的前缀与某个号码相同 就不能被拨打（因为会打到那个与它前缀相同的号码那儿去）。如果存在这种情况 输出NO 否则输出YES
显然 这是字典树的简单运用 注意字典树内存的清理 否则会造成内存泄露！！
*/

#include
#include
#include
using namespace std;


int T,N;
char PhoneNum[10001][11];


typedef struct Dictionary_Tree
{
	//int cou;
	bool flag;   //标记是否是某个单词的结尾 
	struct Dictionary_Tree *next[11];
}DictionaryTree;


DictionaryTree *Free[100005]; 
int Cou_Free;


bool Fun(int k,DictionaryTree &root)  //引用 
{
	DictionaryTree *point=&root,*newnode;
	int t;
	for(int i=0;PhoneNum[k][i]!='\0';i++)
	{
		t=PhoneNum[k][i]-'0';
		//cout<<"@@\n";////
		if(point->next[t] == NULL)
		{
			newnode=(DictionaryTree*)malloc(sizeof(DictionaryTree));
			memset(newnode,0,sizeof(DictionaryTree));
			
			Free[Cou_Free++] = newnode;
			/*newnode->flag=false;
			for(int I=0;I<=10;I++)
			newnode->next[I]=NULL;*/
			
			//newnode->cou=1;
			
			//if(newnode->next[PhoneNum[k][i+1]-'0']==NULL) t=t;///
			
			if(PhoneNum[k][i+1]=='\0') //已到末尾
			{
				newnode->flag=true;
				
			} 
			
			point->next[t]=newnode;  
		}
		
		else
		{
			if(point->next[t]->flag == true || PhoneNum[k][i+1]=='\0') return false;
			//else point->next[t]->cou++;
		}
		
		point=point->next[t];   //指针跟上 
	}
	
	return true;
}


int main()
{
	while(cin>>T)
	{
		while(T--)
		{
			DictionaryTree root={false,{NULL}};
			Cou_Free=0;
			cin>>N;
			for(int i=0;i