HDOJ1238(Substrings) 暴力过

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1258    Accepted Submission(s): 572

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

//1245883 2009-04-08 08:52:09 Accepted 1238 15MS 248K 1234 B C++ Xredman 
#include <iostream>
#include <cstring>
using namespace std;

const int N = 102;

char str[N][N], shortStr[N];
int len[N], shortLen;
int t, n, cc;

bool found(char *pp)
{
    for(int i = 0; i < n; i++)
        if(i != cc && !strstr(str[i], pp) )
            return false;
    return true;
}

int main()
{ 
    char ss[N], rr[N];
    int tt, i, j, k;
    bool flag;
    while(cin>>t)
        while(t-- && cin>>n)
        {
            cc = shortLen = -1;
            for(i = 0; i < n; i++)
            {
                cin>>str[i];
                len[i] = strlen(str[i]);
                if(shortLen == -1 || len[i] < shortLen)
                {
                    cc = i;
                    shortLen = len[i];
                }
            }
            strcpy(shortStr, str[cc]);
            tt = shortLen;flag = false;

            while(tt > 0)
            {
                for(i = 0; i + tt <= shortLen; i++)
                {
                    for(j = i, k = 0; k < tt; k++, j++)
                        ss[k] = shortStr[j];
                    ss[k] = '\0';
                    //cout<<"SS "<<ss<<endl;
                    if(found(ss))
                    {
                        flag = true;
                        break;
                    }
                    for(j = shortLen - i - 1, k = 0; k < tt; k++, j--)
                        rr[k] = shortStr[j];
                    rr[k] = '\0';
                    //cout<<"RR "<<rr<<endl;
                    if(found(rr))
                    {
                        flag = true;
                        break;
                    }
                }
                if(flag)
                    break;
                tt--;
            }
            cout<<tt<<endl;
        }
    return 0;
}