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最新推荐文章于 2025-07-04 09:42:48 发布

原创最新推荐文章于 2025-07-04 09:42:48 发布 · 76 阅读

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该程序旨在判断一组电话号码是否满足一致性条件，即不存在一个电话号码是另一个的前缀。输入包含多个测试用例，每个用例包含若干个电话号码。程序会检查每个用例，并输出“YES”表示号码组一致，或者“NO”表示存在前缀关系。

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:

Emergency 911
Alice 97 625 999
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：找是否存在s,t两个字符串，使s是t的前缀

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int maxn=1e5+10;
int tree[maxn][26];
char s[100];
int vis[maxn];
int t,cnt,f,n,flag;
int insert(char s[])
{
	int l=strlen(s),u=0;
	for(int i=0;i<l;i++)
	{
		int v=s[i]-'0';
		if(!tree[u][v]) tree[u][v]=++cnt;//新加节点 
		else if(i==l-1) flag=1;//到字符串结尾 
		if(vis[u]) flag=1;//如果走到结尾 
		u=tree[u][v];//接下一个节点 
	}
	vis[u]=1;
	return flag;
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(tree,0,sizeof(tree));
		memset(vis,0,sizeof(vis));
		f=0,cnt=0,flag=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%s",s);
			if(insert(s))
				f=1;
		}
		if(f) printf("NO\n");
		else printf("YES\n");
	}
}