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 orchidSample Output
2 2You 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 orchidSample Output
2 2
题目大意：就是给你n个字符串，找出它们的公共最长字符串是多少。
解题思路：看到数据量你就懂了，暴力第一个字符串的每一个子串就可以了，每次都与其他的字符串KMP匹配就行了，需要注意的是对于子串需要反过来判断一下，因为题目说反过来存在也可以。
/*
@Author: Top_Spirit
@Language: C++
*/
#include <bits/stdc++.h>
using namespace std ;
typedef unsigned long long ull ;
typedef long long ll ;
const int Maxn = 100 + 10 ;
const int INF = 0x3f3f3f3f ;
const ull seed = 133 ;
const int MOD = 10007 ;
const double PI = acos(-1.0) ;
int Next[Maxn], slen, plen ,n;
string a[Maxn],ptr,str ;
void get_next (){
Next[0] = -1;
int j = 0, k = -1;
while (j < slen){
if (k == -1 || str[j] == str[k]){
Next[++j] = ++k;
}
else k = Next[k];
}
}
bool kmp (){
int i = 0, j = 0;
while (i < plen){
if (j == -1 || str[j] == ptr[i]){
i++;
j++;
}
else j = Next[j];
if (j == slen) return true;
}
return false ;
}
int main (){
int T;
cin >> T;
while (T--){
cin >> n;
for (int i = 0 ;i < n; i++){
cin >> a[i];
}
int ans = 0;
for (int i = 1; i < a[0].size(); i++){
for (int j = 0 ; j <= a[0].size() - i; j++){
str = a[0].substr(j,i);
// cout << j << " " << i << " " ;
// cout << str << endl ;
slen = str.size();
get_next();
bool flag = 1;
for (int k = 1; k < n; k++){
ptr = a[k];
plen = a[k].size();
if (kmp()) continue ;
reverse(ptr.begin(),ptr.end());
if (kmp()) continue ;
flag = 0;
break;
}
if (flag) ans = max(ans,slen);
}
}
cout << ans << endl ;
}
return 0;
}