HDU1671 Phone List（字典树）

最新推荐文章于 2020-05-09 10:07:54 发布

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本文介绍了一种通过构建前缀树来判断电话号码簿中是否存在号码作为其他号码前缀的问题解决方法。通过输入一系列电话号码并利用前缀树的数据结构进行高效查询与匹配，可以快速确定号码簿是否符合要求。

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

Sample Output

NO
YES

题意: 看每个串的字串是否会与别的串相同，有则输出 NO, else 输出YES

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define N 10

struct node{
  bool flag;
  node *nest[N];
};

node *root;
char phone[10050][20];



void inset(char *c)
{
    node * tem=root;

    while(*c!='\0')
	{
		if(tem->nest[*c-'0']==NULL)
		{
			node * cur;
			cur=(node *)malloc(sizeof(node));

			for(int i=0;i<N;i++)
				cur->nest[i]=NULL;

			cur->flag=false;
			tem->nest[*c-'0']=cur;
		}

		tem=tem->nest[*c-'0'];
		c++;
	}
	tem->flag=true;
}



int seach(char *c)
{
	int i,j;
	node * temp=root;
	
	while(*c!='\0')
	{
		if(temp==NULL||temp->nest[*c-'0']==NULL)
			return 0;

		temp=temp->nest[*c-'0'];
		c++;
	}
	return temp->flag;
}

void delet(node* temp)
{
	if(temp==NULL) return ;

	for(int i=0;i<N;i++)
		if(temp->nest[i]!=NULL)
		 delet(temp->nest[i]);
		 
	free(temp);
}

void inint()
{
	int i,j;
	
	root=(node *) malloc(sizeof(node));
	root->flag=false;

	for(i=0;i<N;i++)
		root->nest[i]=NULL;

}

int main()
{
   int i,j,t,n;
   scanf("%d",&t);
   
   while(t--)
   {
   	  inint();
   	  scanf("%d",&n);
   	  for(i=0;i<n;i++)
		{
			scanf("%s",phone[i]);
			inset(phone[i]);
		}

	    int flag=1;
	    char c[20];

	    for(i=0;i<n;i++)
		{
			if(!flag) break;
		   int len=strlen(phone[i]);
		   
		   memset(c,'\0',sizeof(c));
		

		   for(j=0;j<len-1;j++)
			{
			 c[j]=phone[i][j];
			 
		     if(seach(c))
			 {
			 	flag=0;
			 	break;
			 }
			}
		}

		if(flag)
			printf("YES\n");
		else
			printf("NO\n");

	  delet(root);
   }
   return 0;

}