HDU1671 Phone List 解题报告--字典树

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7202    Accepted Submission(s): 2471

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
  
  

 

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

 

Recommend

lcy

 

#include<iostream>
using namespace std;
#include<cstring>
struct node
{
	int count;
	node *next[10];
	node()
	{
		count=0;
		memset(next,0,sizeof(next));
	}
};
node *root=NULL;
void build(char *s) 
{
	node *p=root; 
	int len=strlen(s);
	int i;
	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->count++;//每个节点记录经过次数
	}
}
int find(char *s) 
{
	node *p=root;
	int len=strlen(s);
	int i;
	for(i=0;i<len;i++)
		p=p->next[s[i]-'0'];
	return p->count;//输出这个节点经过次数
}
void freedom(node *p)//释放空间的函数
{
	for(int i=0;i<10;i++)
	{
		if(p->next[i]!=NULL)
			freedom(p->next[i]);//一旦不为空一直释放下去
	}
	free(p);
}
int main()
{
	int r;
	char a[10001][15];
	scanf("%d",&r);
	while(r--)
	{
		root=new node;
		int n;
		scanf("%d",&n);	
		int i,j;
		for(i=0;i<n;i++)
		{
			scanf("%s",a[i]);
			build(a[i]);
		}
		int f=0;
		for(i=0;i<n;i++)//挨个检索号码的可用
		{			
			if(find(a[i])!=1) {f=0;break;}//一旦经过次数不为1则说明重复直接结束答案为NO
			f=1;
		}
		if(f==1)//所有号码无前段重复，可用
			printf("YES\n");
		else
			printf("NO\n");
		freedom(root);//释放空间
	}
	return 0;
}