http://acm.hdu.edu.cn/showproblem.php?pid=1238

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 7444 Accepted Submission(s): 3358

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

Author

Asia 2002, Tehran (Iran), Preliminary

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求最大公共子序列暴力枚举

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<sstream>
#include<time.h>
#include<utility>
#include<malloc.h>
#include<stdexcept>

using namespace std;

char p[110][110];

int main ()
{
    int t,m,l;
    cin>>t;
    while ( t-- )
    {
        int n;
        cin>>n;
        m = 1000;
        for(int i=0;i<n;i++)
            {
                cin>>p[i];
                 if ( m > strlen (p[i]))
                 {
                     m = strlen(p[i]);
                     l = i;
                 }
            }

        char os[110],ps[110];
        int ans = 0;

        for(int i=0;i<strlen(p[l]);i++)
            {
                for(int  j= i;j<strlen(p[l]);j++ )
                {
                    int b = 0;
                    int c = j-i;
                    for(int k = i;k<=j;k++)
                    {
                        os[b++] = p[l][k];
                        ps[c--] = p[l][k]; 
                    }

                    os[b] = '\0';
                    ps[j-i+1] = '\0';

                    int ok =1;

                    for(int k =0;k<n;k++)
                    {
                         if (!strstr(p[k],os) && !strstr (p[k],ps))
                         {
                             ok = 0;
                             break;
                         }
                    }

                    if (ok && ans < strlen(ps))
                    {
                        ans = strlen(ps);
                    }
                }
            }
        cout<<ans<<endl;
    }
    return 0;
}