HDOJ1238(Substrings)

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1268    Accepted Submission(s): 577

Problem Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

 

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

 

Output

There should be one line per test case containing the length of the largest string found.

 

Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

 

Sample Output

2
2

 

 

//1258362 2009-04-11 13:03:15 Accepted 1238 15MS 248K 1259 B C++ Xredman 
#include <iostream>
#include <cstring>
using namespace std;

const int N = 105;

typedef char STR[N];

STR str[N];
int sti, n;

bool find(STR substr)
{
    int i;
    for(i = 0; i < n; i++)
        if(i != sti && !strstr(str[i], substr))
            return false;
    return true;
}

int main()
{
    int T, i, j, minlen, kk , cc;
    bool mark;
    STR substr;
    cin>>T;
    while(T-- && cin>>n)
    {
        sti = 0;minlen = -1;
        for(i = 0; i < n; i++)
        {
            cin>>str[i];
            if(minlen == -1 || strlen(str[i]) < minlen)
            {
                minlen = strlen(str[i]);;
                sti = i;
            }
        }
        kk = minlen;mark = false;
        while(kk)
        {
            for(i = 0; i + kk <= minlen; i++)
            {
                for(cc = 0, j = i; cc < kk; j++, cc++)
                substr[cc] = str[sti][j];    
                substr[cc] = '\0';
                //cout<<"substr: "<<substr<<endl;
                if(find(substr))
                {
                    mark = true;
                    break;
                }
            }
            if(mark)
                break;
            for(i = minlen - 1; i - kk >= -1; i--)
            {
                for(cc = 0, j = i; cc < kk; j--, cc++)
                    substr[cc] = str[sti][j];
                substr[cc] = '\0';
                //cout<<"substr2: "<<substr<<endl;
                if(find(substr))
                {
                    mark = true;
                    break;
                }
            }
            if(mark)
                break;
            kk--;
        }
        cout<<kk<<endl;
    }

}