山东省第一届ACM省赛 A SDUT 2151 Phone Number（暴力）

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本文介绍了一种用于判断电话号码列表中是否存在前缀冲突的算法。通过逐个比较电话号码，该算法能够确定是否有一个号码是另一个号码的前缀，从而避免在实际呼叫过程中发生错误。文章提供了完整的代码实现及示例。

Phone Number

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

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

示例输出

NO
YES

#include <bits/stdc++.h>
using namespace std;
char str[1010][100100];
int main()
{
    int n;
    while(~scanf("%d",&n) && n)
    {
        bool flag = false;
        for(int i=0; i<n; i++)
        {
            scanf("%s",str[i]);
            if(!flag)
            {
                for(int j=0; j<i; j++)
                {
                    bool f = 0;
                    for(int e  = 0;  str[i][e] && str[j][e]; e++)
                    {
                        if(str[i][e] != str[j][e])
                        {
                            f = 1;
                            break;
                        }
                    }
                    if(!f)
                    {
                        flag = 1;
                        break;
                    }
                }
            }
        }
        if(flag)
            cout<<"NO"<<endl;
        else
            cout<<"YES"<<endl;
    }
}