POJ 1226 Substrings

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

Tehran 2002 Preliminary

 

 

解题：霸蛮好了。。。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 string str[];
 int main() {
     int t,i,j,k,n;
     bool  flag;
     scanf("%d",&t);
     while(t--){
         scanf("%d",&n);
         for(i = ; i < n; i++)
             cin>>str[i];
         sort(str,str+n);
         flag = false;
         for(k = str[].length(); k; k--){
             for(i = ; i + k <= str[].length(); i++){
                 string a = str[].substr(i,k);
                 string b(a.rbegin(),a.rend());
                 for(j = ; j < n; j++)
                     if(str[j].find(a) == - && str[j].find(b) == -) break;
                 if(j == n) {flag = true;break;}
             }
             if(flag) break;
         }
         flag?printf("%d\n",k):puts("");
     }
     return ;
 }