Substrings
Time Limit: 1000ms
Memory Limit: 10000KB
This problem will be judged on PKU. Original ID: 1226
64-bit integer IO format: %lld Java class name: Main
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 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
Source
解题：霸蛮好了。。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 string str[110];
18 int main() {
19 int t,i,j,k,n;
20 bool flag;
21 scanf("%d",&t);
22 while(t--){
23 scanf("%d",&n);
24 for(i = 0; i < n; i++)
25 cin>>str[i];
26 sort(str,str+n);
27 flag = false;
28 for(k = str[0].length(); k; k--){
29 for(i = 0; i + k <= str[0].length(); i++){
30 string a = str[0].substr(i,k);
31 string b(a.rbegin(),a.rend());
32 for(j = 1; j < n; j++)
33 if(str[j].find(a) == -1 && str[j].find(b) == -1) break;
34 if(j == n) {flag = true;break;}
35 }
36 if(flag) break;
37 }
38 flag?printf("%d\n",k):puts("0");
39 }
40 return 0;
41 }