Phone Number 2010年山东省第一届ACM大学生程序设计竞赛

Phone Number

Time Limit: 1000MS Memory limit: 65536K

题目描述

We know that if a phone number A is another phone number B’s prefix, B is not able to be called. For an example, A is 123 while B is 12345, after pressing 123, we call A, and not able to call B.
Given N phone numbers, your task is to find whether there exits two numbers A and B that A is B’s prefix.

输入

 The input consists of several test cases.
 The first line of input in each test case contains one integer N (0<N<1001), represent the number of phone numbers.
 The next line contains N integers, describing the phone numbers.
 The last case is followed by a line containing one zero.

输出

 For each test case, if there exits a phone number that cannot be called, print “NO”, otherwise print “YES” instead.

示例输入

2
012
012345
2
12
012345
0

示例输出

NO
YES

#include<iostream>
#include<string>
using namespace std;
string a[1005],k;

int b[1005];
int main()
{
	int n,i,j,p,s;
	while(cin>>n&&n)
	{
		s=0;
		for(i=0;i<n;i++)
		cin>>a[i];
			
	
		for(i=0;i<n-1;i++)
			for(j=0;j<n-1-i;j++)
			{if(a[j]>a[j+1])
			{k=a[j];
			a[j]=a[j+1];
			a[j+1]=k;}}
			//for(i=0;i<n;i++)
				//cout<<a[i]<<endl;
			for(i=0;i<n;i++)
				b[i]=a[i].size();
			for(i=0;i<n-1;i++)
			{p=b[i]<b[i+1]?b[i]:b[i+1];
				for(j=0;j<p;j++)
					 if(a[i][j]!=a[i+1][j]) break;
					 else if(j==p-1)
						 s=1;
			}
		//	for(i=0;i<n-1;i++)
			//	if(a[i]==a[i+1])
				//	s=2;
			if(s==1)
				cout<<"NO"<<endl;
			else
				if(s==0)
					cout<<"YES"<<endl;
			
				
	}

return 0;
}