ZOJ Problem Set - 2876 Phone List

最新推荐文章于 2018-07-25 16:04:54 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问题，确保任一号码不是其他号码的前缀。通过排序和比较相邻号码的方法实现，提供了完整的AC代码。

Phone List

Time Limit: 5 Seconds Memory Limit: 32768 KB

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

2
3
911
97625999
91125426
5
113
12340
123440
12345
98346

Sample Output

NO
YES

Source: The 2007 Nordic Collegiate Programming Contest

分析：

题意：

给定若干电话号码（每个电话号码的长度不超过10个数字），要求判断是否存在一个电话号码是其他某个电话号码的前缀，如果存在，输出“no”；否则输出“yes”。

把电话号码当做字符串处理。先对所有电话号码排序（从小到大），这样只要判断某个字符串是否是它的下一个字符串的前缀就行。考虑全部字符串，从左到右扫描一遍，如果都不满足前缀的条件，则输出“yes”；如果中途有一个满足，输出“no”，并break。

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
string s;
char c[15];
vector<string> v;
int main()
{
int t,n,i,j;
scanf("%d",&t);
while(t--)
{
v.clear();//注意每次清空
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s",c);
s=c;
v.push_back(s);
}
sort(v.begin(),v.end());
for(i=0;i<n-1;i++)
{
if(v[i+1].find(v[i])==0)
{
printf("NO\n");
break;
}
}
if(i==n-1)
printf("YES\n");
}
return 0;
}