hdu 1671 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


本题的题意是,给你n个串的电话号码。问有没有可能一个电话号码是其他的 号码的前缀,、
比如911 它就是91156的前缀。如果有的话输出NO。没有则输出YES
这题的解法就是字典树,首先套用字典树的模板,建立字典树,注意清空root
然后将输入的电话号码存入字典树里面,然后寻找遍历字典树,如果遍历完一个电话号码
之后,看后面的节点是否为空,如果不为空,说明有其他的电话号码是以这个前缀的,标记,
然后判断输出;,,而且注意一点就是结束之后记得清空,不然很容易mle。具体实现如下:

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
struct node
{
	struct node *next[10];
};

struct node *root;
void charu(char *qq)
{
	struct node *now1,*now2;
	int i,j;
	int l=strlen(qq);
	now1=root;
	
	for(i=0;i<l;i++)
	{
		int k=qq[i]-'0';
		if(now1->next[k]) now1=now1->next[k];
		else 
		{
			now2=new struct node;
			for(j=0;j<10;j++) now2->next[j]=0;
			now1->next[k]=now2;
			now1=now2;
		}
	}
}
int find(char *s)
{
	int i;
	struct node *now;
	now=root;
	int flag;
	int l=strlen(s);
	
	for(i=0;i<l;i++) now=now->next[s[i]-'0'];
	flag=0;
	for(i=0;i<10;i++) 
	if(now->next[i])
	{
		flag=1; break;
	}

}
void qing(struct node *pp)
{
	int i;
	for(i=0;i<10;i++)
	if(pp->next[i])
	qing(pp->next[i]);
	delete(pp);
}

int main()
{
	int bbs,n,i,j,ans;
	char ss[10010][12];
	cin>>bbs;
	while(bbs--)
	{
		cin>>n;
		root=new struct node;
		for(i=0;i<10;i++) root->next[i]=0;
		
		for(i=0;i<n;i++)
		{
			scanf("%s",ss[i]);
			charu(ss[i]);
		}
		ans=0;
		for(i=0;i<n;i++)
		{
			ans=find(ss[i]);
			if(ans)  break;
		}
		if(ans)  cout<<"NO"<<endl;
		else cout<<"YES"<<endl;
		qing(root);
	}
	return 0;
	
}


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