UVALive 4423 String LD 暴力

A - String LD

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status Practice UVALive 4423

Description

Stringld(left delete) is a function that gets a string and deletes its leftmost character (for instance Stringld(``acm") returns ``cm").

You are given a list of distinct words, and at each step, we apply stringld on every word in the list. Write a program that determines the number of steps that can be applied until at least one of the conditions become true:

 

1. A word becomes empty string, or
2. a duplicate word is generated.

For example, having the list of words aab, abac, and caac, applying the function on the input for the first time results in ab, bac, and aac. For the second time, we get b, ac, and ac. Since in the second step, we have two ac strings, the condition 2 is true, and the output of your program should be 1. Note that we do not count the last step that has resulted in duplicate string. More examples are found in the sample input and output section.

Input

There are multiple test cases in the input. The first line of each test case is n(1n100) , the number of words.

Each of the next n lines contains a string of at most 100 lower case characters.

The input terminates with a line containing `0'.

Output

For each test case, write a single line containing the maximum number of stringld we can call.

Sample Input

4 
aaba 
aaca 
baabcd 
dcba 
3 
aaa 
bbbb 
ccccc 
0

Sample Output

1
2

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100001
const int inf=0x7fffffff;   //无限大

//string &erase(int pos = 0, int n = npos);//删除pos开始的n个字符，返回修改后的字符串

string s[1001];
int main()
{
    int n;
    while(cin>>n)
    {
        if(n==0)
            break;
        for(int i=0;i<n;i++)
            cin>>s[i];
        int flag=0;
        int flag2=0;
        if(n!=1)
        while(1)
        {
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                {
                    if(i==j)
                        continue;
                    for(int k=0;k<s[i].size();k++)
                    {
                        if(s[i][k]!='(')
                            break;
                        if(k==s[i].size()-1)
                        {
                            flag2=1;
                            break;
                        }
                    }
                    if(s[i]==s[j])
                    {
                        flag2=1;
                        break;
                    }
                }
            }
            if(flag2==1)
                break;
            for(int i=0;i<n;i++)
            {
                s[i][flag]='(';
            }
            flag++;
        }
        flag--;
        if(flag<0)
            flag=0;
        cout<<flag<<endl;

    }
    return 0;
}