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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#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;
}
