POJ 题目3630 Phone List（字符串，水）

最新推荐文章于 2020-10-04 21:18:14 发布

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最新推荐文章于 2020-10-04 21:18:14 发布

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本文介绍了一个电话号码列表的一致性检查算法。该算法确保列表中任一电话号码都不是另一个号码的前缀，避免了拨打时可能产生的误接情况。通过输入一系列电话号码，并对这些号码进行比较判断，最终输出列表是否一致。

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Phone List

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23729		Accepted: 7293

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:

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

Source

Nordic 2007

ac代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define min(a,b) (a>b?b:a)
struct str
{
	char s[20];
}b[10010];
int cmp(const void *a,const void *b)
{
	return strcmp((*(struct str *)a).s,(*(struct str *)b).s);
}
int n;
int jud()
{
	int i,j,k;
	for(i=0;i<n-1;i++)
	{
		int len=strlen(b[i].s);
		for(j=0;j<len;j++)
		{
			if(b[i].s[j]!=b[i+1].s[j])
				break;
		}
		if(j==len)
			return 1;
	}
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		//int n;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%s",b[i].s);
		}
		qsort(b,n,sizeof(b[0]),cmp);
		int ans=jud();
		if(ans)
			printf("NO\n");
		else
			printf("YES\n");
	}
}